Thu. Jul 18th, 2024

Jul 31, 2020

# Q1. (a)  $\fn_cm&space;{\color{Blue}&space;m+\frac{1}{m}=\sqrt{3}}$  হলে,

## (ii) $\fn_cm&space;{\color{Blue}&space;{{m}^{3}}+\frac{1}{{{m}^{3}}}}$

এদের সরলতম মান নির্ণয় করি।

সমাধান (i) :

(i) $\fn_cm&space;{{m}^{2}}+\frac{1}{{{m}^{2}}}$

$\fn_cm&space;=\left&space;(&space;m+\frac{1}{m}&space;\right&space;)^{2}-2\times&space;m\times&space;\frac{1}{m}$

$\fn_cm&space;=\left&space;(&space;\sqrt{3}&space;\right&space;)^{2}-2\:&space;\:&space;{\color{Blue}&space;\left&space;[&space;\because&space;m+\frac{1}{m}=\sqrt{3}&space;\right&space;]}$

$\fn_cm&space;=3-2$

$\fn_cm&space;=1$

উত্তরঃ নির্ণেয় $\fn_cm&space;{\color{DarkGreen}&space;{{m}^{2}}+\frac{1}{{{m}^{2}}}}$ এর সরলতম মান  1

সমাধান (ii) :

(ii) $\fn_cm&space;m^{3}+\frac{1}{m^{3}}$

$\fn_cm&space;=\left&space;(&space;m+\frac{1}{m}&space;\right&space;)^{3}-3\times&space;m\times&space;\frac{1}{m}\left&space;(&space;m+\frac{1}{m}&space;\right&space;)$

$\fn_cm&space;=\left&space;(&space;\sqrt{3}&space;\right&space;)^{3}-3\times&space;\sqrt{3}\:&space;\:&space;{\color{Blue}&space;\left&space;[&space;\because&space;m+\frac{1}{m}=\sqrt{3}&space;\right&space;]}$

$\fn_cm&space;=3\sqrt{3}-3\sqrt{3}$

$\fn_cm&space;=0$

উত্তরঃ নির্ণেয় $\fn_cm&space;{\color{DarkGreen}&space;{{m}^{3}}+\frac{1}{{{m}^{3}}}}$ এর সরলতম মান  0

 Q1. (b) দেখাই যে, $\fn_cm&space;\small&space;{\color{Blue}&space;\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=2\sqrt{15}}$ সমাধানঃ বামপক্ষ $\fn_cm&space;\small&space;=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$ $\fn_cm&space;\small&space;=\frac{\left&space;(&space;\sqrt{5}+\sqrt{3}&space;\right&space;)^{2}-\left&space;(&space;\sqrt{5}-\sqrt{3}&space;\right&space;)^{2}}{\left&space;(&space;\sqrt{5}-\sqrt{3}&space;\right&space;)\left&space;(&space;\sqrt{5}+\sqrt{3}&space;\right&space;)}$ $\small&space;=\frac{\left&space;[\left&space;(&space;\sqrt{5}&space;\right&space;)^{2}+2\times&space;\sqrt{5}\times&space;\sqrt{3}+\left&space;(&space;\sqrt{3}&space;\right&space;)^{2}&space;\right&space;]-\left&space;[\left&space;(&space;\sqrt{5}&space;\right&space;)^{2}-2\times&space;\sqrt{5}\times&space;\sqrt{3}+\left&space;(&space;\sqrt{3}&space;\right&space;)^{2}&space;\right&space;]}{\left&space;(&space;\sqrt{5}&space;\right&space;)^{2}-\left&space;(&space;\sqrt{3}&space;\right&space;)^{2}}$$\fn_cm&space;{\color{Blue}&space;\left&space;[&space;\because&space;\left&space;(&space;a+b&space;\right&space;)\left&space;(&space;a-b&space;\right&space;)=a^{2}-b^{2}&space;\right&space;]}$ $\fn_cm&space;=\frac{5+2\sqrt{15}+3-5+2\sqrt{15}-3}{5-3}$ $\fn_cm&space;=\frac{4\sqrt{15}}{2}$ $\fn_cm&space;=2\sqrt{15}=$ ডানপক্ষ ( প্রমাণিত )

 Q2. সরল করি : (a) $\fn_cm&space;{\color{Blue}&space;\frac{\sqrt{2}\left(&space;2+\sqrt{3}&space;\right)}{\sqrt{3}\left(&space;\sqrt{3}+1&space;\right)}-\frac{\sqrt{2}\left(&space;2-\sqrt{3}&space;\right)}{\sqrt{3}\left(&space;\sqrt{3}-1&space;\right)}}$ সমাধানঃ  $\fn_cm&space;\frac{\sqrt{2}\left(&space;2+\sqrt{3}&space;\right)}{\sqrt{3}\left(&space;\sqrt{3}+1&space;\right)}-\frac{\sqrt{2}\left(&space;2-\sqrt{3}&space;\right)}{\sqrt{3}\left(&space;\sqrt{3}-1&space;\right)}$ $\fn_cm&space;=\frac{\sqrt{2}}{\sqrt{3}}\left&space;[\frac{\left(&space;2+\sqrt{3}&space;\right)}{\left(&space;\sqrt{3}+1&space;\right)}-\frac{\left(&space;2-\sqrt{3}&space;\right)}{\left(&space;\sqrt{3}-1&space;\right)}&space;\right&space;]$ $\fn_cm&space;=\frac{\sqrt{2}}{\sqrt{3}}\left&space;[\frac{\left&space;(&space;2+\sqrt{3}&space;\right&space;)\left&space;(&space;\sqrt{3}-1&space;\right&space;)-\left&space;(&space;2-\sqrt{3}&space;\right&space;)\left&space;(&space;\sqrt{3}+1&space;\right&space;)}{\left&space;(&space;\sqrt{3}+1&space;\right&space;)\left&space;(&space;\sqrt{3}-1&space;\right&space;)}&space;\right&space;]$ $\fn_cm&space;=\frac{\sqrt{2}}{\sqrt{3}}\left&space;[&space;\frac{\left&space;\{&space;2&space;\sqrt{3}-2+3-\sqrt{3}\right&space;\}-\left&space;\{&space;2&space;\sqrt{3}+2-3-\sqrt{3}&space;\right&space;\}}{\left&space;(&space;\sqrt{3}&space;\right&space;)^{2}-\left&space;(&space;1&space;\right&space;)^{2}}&space;\right&space;]\;&space;\;&space;{\color{Blue}&space;\left&space;[&space;\because&space;\left&space;(&space;a+b&space;\right&space;)\left&space;(&space;a-b&space;\right&space;)=a^{2}-b^{2}&space;\right&space;]}$ $\fn_cm&space;=\frac{\sqrt{2}}{\sqrt{3}}\left&space;[&space;\frac{2\sqrt{3}-2+3-\sqrt{3}-2\sqrt{3}-2+3+\sqrt{3}}{3-1}&space;\right&space;]$ $\fn_phv&space;=\frac{\sqrt{2}}{\sqrt{3}}\left&space;[&space;\frac{2}{2}&space;\right&space;]$ $\fn_cm&space;=\frac{\sqrt{2}}{\sqrt{3}}$ $\fn_cm&space;=\frac{\sqrt{2}\times&space;\sqrt{3}}{\sqrt{3}\times&space;\sqrt{3}}$ [ হরের করণী নিরসন করে পাই ] $\fn_cm&space;=\frac{\sqrt{6}}{3}$ উত্তরঃ নির্ণেয় সরলতম মান $\fn_cm&space;{\color{DarkGreen}&space;\frac{\sqrt{6}}{3}}$

 Q2. সরল করি : (b) $\fn_cm&space;{\color{Blue}&space;\frac{3\sqrt{7}}{\sqrt{5}+\sqrt{2}}-\frac{5\sqrt{5}}{\sqrt{2}+\sqrt{7}}+\frac{2\sqrt{2}}{\sqrt{7}+\sqrt{5}}}$ সমাধানঃ  $\fn_cm&space;\frac{3\sqrt{7}}{\sqrt{5}+\sqrt{2}}-\frac{5\sqrt{5}}{\sqrt{2}+\sqrt{7}}+\frac{2\sqrt{2}}{\sqrt{7}+\sqrt{5}}$ $\fn_cm&space;=\frac{3\sqrt{7}\left&space;(&space;\sqrt{5}-\sqrt{2}&space;\right&space;)}{\left&space;(&space;\sqrt{5}+\sqrt{2}&space;\right&space;)\left&space;(&space;\sqrt{5}-\sqrt{2}&space;\right&space;)}-\frac{5\sqrt{5}\left&space;(&space;\sqrt{2}-\sqrt{7}&space;\right&space;)}{\left&space;(&space;\sqrt{2}+\sqrt{7}&space;\right&space;)\left&space;(&space;\sqrt{2}-\sqrt{7}&space;\right&space;)}+\frac{2\sqrt{2}\left&space;(&space;\sqrt{7}-\sqrt{5}&space;\right&space;)}{\left&space;(&space;\sqrt{7}+\sqrt{5}&space;\right&space;)\left&space;(&space;\sqrt{7}-\sqrt{5}&space;\right&space;)}$ [হরের করণী নিরসন করে পাই ] $\fn_cm&space;=\frac{3\sqrt{7}\left&space;(&space;\sqrt{5}-\sqrt{2}&space;\right&space;)}{\left&space;(&space;\sqrt{5}&space;\right&space;)^{2}-\left&space;(&space;\sqrt{2}&space;\right&space;)^{2}}-\frac{5\sqrt{5}\left&space;(&space;\sqrt{2}-\sqrt{7}&space;\right&space;)}{\left&space;(&space;\sqrt{2}&space;\right&space;)^{2}-\left&space;(&space;\sqrt{7}&space;\right&space;)^{2}}+\frac{2\sqrt{2}\left&space;(&space;\sqrt{7}-\sqrt{5}&space;\right&space;)}{\left&space;(&space;\sqrt{7}&space;\right&space;)^{2}-\left&space;(&space;\sqrt{5}&space;\right&space;)^{2}}$ $=\frac{3\sqrt{7}\left&space;(&space;\sqrt{5}-\sqrt{2}&space;\right&space;)}{5-2}-\frac{5\sqrt{5}\left&space;(&space;\sqrt{2}-\sqrt{7}&space;\right&space;)}{2-7}+\frac{2\sqrt{2}\left&space;(&space;\sqrt{7}-\sqrt{5}&space;\right&space;)}{7-5}$ $=\frac{3\sqrt{7}\left&space;(&space;\sqrt{5}-\sqrt{2}&space;\right&space;)}{3}-\frac{5\sqrt{5}\left&space;(&space;\sqrt{2}-\sqrt{7}&space;\right&space;)}{-5}+\frac{2\sqrt{2}\left&space;(&space;\sqrt{7}-\sqrt{5}&space;\right&space;)}{2}$ $\fn_cm&space;=\sqrt{7}\left&space;(&space;\sqrt{5}-\sqrt{2}&space;\right&space;)+\sqrt{5}\left&space;(&space;\sqrt{2}-\sqrt{7}&space;\right&space;)+\sqrt{2}\left&space;(&space;\sqrt{7}-\sqrt{5}&space;\right&space;)$ $\fn_cm&space;=\sqrt{35}-\sqrt{14}+\sqrt{10}-\sqrt{35}+\sqrt{14}-\sqrt{10}$ $\fn_cm&space;=0$ উত্তরঃ নির্ণেয় সরলতম মান $\fn_jvn&space;{\color{DarkGreen}&space;0}$

 Q2. সরল করি : (c) $\fn_cm&space;{\color{Blue}&space;\frac{4\sqrt{3}}{2-\sqrt{2}}-\frac{30}{4\sqrt{3}-\sqrt{18}}-\frac{\sqrt{18}}{3-\sqrt{12}}}$ সমাধানঃ  $\fn_cm&space;\frac{4\sqrt{3}}{2-\sqrt{2}}-\frac{30}{4\sqrt{3}-\sqrt{18}}-\frac{\sqrt{18}}{3-\sqrt{12}}$ $\fn_cm&space;=\frac{4\sqrt{3}\left&space;(&space;2+\sqrt{2}&space;\right&space;)}{\left&space;(&space;2-\sqrt{2}&space;\right&space;)\left&space;(&space;2+\sqrt{2}&space;\right&space;)}-\frac{30\left&space;(&space;4\sqrt{3}+\sqrt{18}&space;\right&space;)}{\left&space;(&space;4\sqrt{3}-\sqrt{18}&space;\right&space;)\left&space;(&space;4\sqrt{3}+\sqrt{18}&space;\right&space;)}-\frac{\sqrt{18}\left&space;(&space;3+\sqrt{12}&space;\right&space;)}{\left&space;(&space;3-\sqrt{12}&space;\right&space;)\left&space;(&space;3+\sqrt{12}&space;\right&space;)}$  [ হরের করণী নিরসন করে পাই ] $\fn_cm&space;=\frac{4\sqrt{3}\left&space;(&space;2+\sqrt{2}&space;\right&space;)}{\left&space;(&space;2&space;\right&space;)^{2}-\left&space;(&space;\sqrt{2}&space;\right&space;)^{2}}-\frac{30\left&space;(&space;4&space;\sqrt{3}+\sqrt{18}&space;\right&space;)}{\left&space;(&space;4\sqrt{3}&space;\right&space;)^{2}-\left&space;(&space;\sqrt{18}&space;\right&space;)^{2}}-\frac{\sqrt{18}\left&space;(&space;3+\sqrt{12}&space;\right&space;)}{\left&space;(&space;3&space;\right&space;)^{2}-\left&space;(&space;\sqrt{12}&space;\right&space;)^{2}}\;&space;{\color{Blue}&space;\left&space;[&space;\because&space;\left&space;(&space;a+b&space;\right&space;)\left&space;(&space;a-b&space;\right&space;)=a^{2}-b^{2}&space;\right&space;]}$ $\fn_cm&space;=\frac{4\sqrt{3}\left&space;(&space;2+\sqrt{2}&space;\right&space;)}{4-2}-\frac{30\left&space;(&space;4&space;\sqrt{3}+\sqrt{18}&space;\right&space;)}{48-18}-\frac{\sqrt{18}\left&space;(&space;3+\sqrt{12}&space;\right&space;)}{9-12}$ $\fn_cm&space;=\frac{4\sqrt{3}\left&space;(&space;2+\sqrt{2}&space;\right&space;)}{2}-\frac{30\left&space;(&space;4\sqrt{3}+3\sqrt{2}&space;\right&space;)}{30}-\frac{3\sqrt{2}\left&space;(&space;3+2\sqrt{3}&space;\right&space;)}{-3}\;&space;{\color{Blue}&space;\left&space;[&space;\because&space;\sqrt{18}=3\sqrt{2}&space;,\;&space;\sqrt{12}=2\sqrt{3}\right&space;]}$ $\fn_cm&space;=2\sqrt{3}\left&space;(&space;2+\sqrt{2}&space;\right&space;)-\left&space;(&space;4\sqrt{3}+3\sqrt{2}&space;\right&space;)+\sqrt{2}\left&space;(&space;3+2\sqrt{3}&space;\right&space;)$ $\fn_cm&space;=4\sqrt{6}+2\sqrt{6}-4\sqrt{3}-3\sqrt{2}+3\sqrt{2}+2\sqrt{6}$ $\fn_cm&space;=4\sqrt{6}$ উত্তরঃ নির্ণেয় সরলতম মান $\fn_cm&space;{\color{DarkGreen}&space;4\sqrt{6}}$

Koshe dekhi 9.3 class 10

 Q2. সরল করি : (d) $\fn_cm&space;{\color{Blue}&space;\frac{3\sqrt{2}}{\sqrt{3}+\sqrt{6}}-\frac{4\sqrt{3}}{\sqrt{6}+\sqrt{2}}+\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}}$ সমাধানঃ  $\fn_cm&space;\frac{3\sqrt{2}}{\sqrt{3}+\sqrt{6}}-\frac{4\sqrt{3}}{\sqrt{6}+\sqrt{2}}+\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}$ $\fn_cm&space;=\frac{3\sqrt{2}\left&space;(\sqrt{3}-\sqrt{6}&space;\right&space;)}{\left&space;(\sqrt{3}+\sqrt{6}&space;\right&space;)\left&space;(\sqrt{3}-\sqrt{6}&space;\right&space;)}-\frac{4\sqrt{3}\left&space;(\sqrt{6}-\sqrt{2}&space;\right&space;)}{\left&space;(\sqrt{6}+\sqrt{2}&space;\right&space;)\left&space;(\sqrt{6}-\sqrt{2}&space;\right&space;)}+\frac{\sqrt{6}\left&space;(\sqrt{2}-\sqrt{3}&space;\right&space;)}{\left&space;(\sqrt{2}+\sqrt{3}&space;\right&space;)\left&space;(\sqrt{2}-\sqrt{3}&space;\right&space;)}$ [ হরের করণী নিরসন করে পাই ] $\fn_cm&space;=\frac{3\sqrt{2}\left&space;(\sqrt{3}-\sqrt{6}&space;\right&space;)}{\left&space;(&space;\sqrt{3}&space;\right&space;)^{2}-\left&space;(&space;\sqrt{6}&space;\right&space;)^{2}}-\frac{4\sqrt{3}\left&space;(&space;\sqrt{6}-\sqrt{2}&space;\right&space;)}{\left&space;(&space;\sqrt{6}&space;\right&space;)^{2}-\left&space;(&space;\sqrt{2}&space;\right&space;)^{2}}+\frac{\sqrt{6}\left&space;(&space;\sqrt{2}-\sqrt{3}&space;\right&space;)}{\left&space;(&space;\sqrt{2}&space;\right&space;)^{2}-\left&space;(&space;\sqrt{3}&space;\right&space;)^{2}}\;&space;{\color{Blue}&space;\left&space;[&space;\because&space;\left&space;(&space;a+b&space;\right&space;)\left&space;(&space;a-b&space;\right&space;)=a^{2}-b^{2}&space;\right&space;]}$ $\fn_cm&space;=\frac{3\sqrt{2}\left&space;(\sqrt{3}-\sqrt{6}&space;\right&space;)}{3-6}-\frac{4\sqrt{3}\left&space;(&space;\sqrt{6}-\sqrt{2}&space;\right&space;)}{6-2}+\frac{\sqrt{6}\left&space;(&space;\sqrt{2}-\sqrt{3}&space;\right&space;)}{2-3}$ $\fn_cm&space;=\frac{3\sqrt{2}\left&space;(\sqrt{3}-\sqrt{6}&space;\right&space;)}{-3}-\frac{4\sqrt{3}\left&space;(&space;\sqrt{6}-\sqrt{2}&space;\right&space;)}{4}+\frac{\sqrt{6}\left&space;(&space;\sqrt{2}-\sqrt{3}&space;\right&space;)}{-1}$ $\fn_cm&space;=-\sqrt{2}\left&space;(&space;\sqrt{3}-\sqrt{6}&space;\right&space;)-\sqrt{3}\left&space;(&space;\sqrt{6}-\sqrt{2}&space;\right&space;)-\sqrt{6}\left&space;(&space;\sqrt{2}-\sqrt{3}&space;\right&space;)$ $\fn_cm&space;=-\sqrt{6}+\sqrt{12}-\sqrt{18}+\sqrt{6}-\sqrt{12}+\sqrt{18}$ $\fn_cm&space;=0$ উত্তরঃ নির্ণেয় সরলতম মান 0

Koshe dekhi 9.3 class 10

 Q3. যদি  $\fn_cm&space;{\color{Blue}&space;x=2,y=3}$  এবং   $\fn_cm&space;{\color{Blue}&space;z=6}$  হয়, তবে,  $\small&space;{\color{Blue}&space;\frac{3\sqrt{x}}{\sqrt{y}+\sqrt{z}}-\frac{4\sqrt{y}}{\sqrt{z}+\sqrt{x}}+\frac{\sqrt{z}}{\sqrt{x}+\sqrt{y}}}$  এর মান হিসাব করে লিখি। সমাধানঃ  $\fn_cm&space;\frac{3\sqrt{x}}{\sqrt{y}+\sqrt{z}}-\frac{4\sqrt{y}}{\sqrt{z}+\sqrt{x}}+\frac{\sqrt{z}}{\sqrt{x}+\sqrt{y}}$ $\fn_cm&space;=\frac{3\sqrt{2}}{\sqrt{3}+\sqrt{6}}-\frac{4\sqrt{3}}{\sqrt{6}+\sqrt{2}}+\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}$ [ x , y ,z  এর মান বসিয়ে পাই ] $\fn_cm&space;=\frac{3\sqrt{2}\left&space;(\sqrt{3}-\sqrt{6}&space;\right&space;)}{\left&space;(\sqrt{3}+\sqrt{6}&space;\right&space;)\left&space;(\sqrt{3}-\sqrt{6}&space;\right&space;)}-\frac{4\sqrt{3}\left&space;(\sqrt{6}-\sqrt{2}&space;\right&space;)}{\left&space;(\sqrt{6}+\sqrt{2}&space;\right&space;)\left&space;(\sqrt{6}-\sqrt{2}&space;\right&space;)}+\frac{\sqrt{6}\left&space;(\sqrt{2}-\sqrt{3}&space;\right&space;)}{\left&space;(\sqrt{2}+\sqrt{3}&space;\right&space;)\left&space;(\sqrt{2}-\sqrt{3}&space;\right&space;)}$ [ হরের করণী নিরসন করে পাই ] $\fn_cm&space;=\frac{3\sqrt{2}\left&space;(\sqrt{3}-\sqrt{6}&space;\right&space;)}{\left&space;(&space;\sqrt{3}&space;\right&space;)^{2}-\left&space;(&space;\sqrt{6}&space;\right&space;)^{2}}-\frac{4\sqrt{3}\left&space;(&space;\sqrt{6}-\sqrt{2}&space;\right&space;)}{\left&space;(&space;\sqrt{6}&space;\right&space;)^{2}-\left&space;(&space;\sqrt{2}&space;\right&space;)^{2}}+\frac{\sqrt{6}\left&space;(&space;\sqrt{2}-\sqrt{3}&space;\right&space;)}{\left&space;(&space;\sqrt{2}&space;\right&space;)^{2}-\left&space;(&space;\sqrt{3}&space;\right&space;)^{2}}\;&space;{\color{Blue}&space;\left&space;[&space;\because&space;\left&space;(&space;a+b&space;\right&space;)\left&space;(&space;a-b&space;\right&space;)=a^{2}-b^{2}&space;\right&space;]}$ $\fn_cm&space;=\frac{3\sqrt{2}\left&space;(\sqrt{3}-\sqrt{6}&space;\right&space;)}{3-6}-\frac{4\sqrt{3}\left&space;(&space;\sqrt{6}-\sqrt{2}&space;\right&space;)}{6-2}+\frac{\sqrt{6}\left&space;(&space;\sqrt{2}-\sqrt{3}&space;\right&space;)}{2-3}$ $\fn_cm&space;=\frac{3\sqrt{2}\left&space;(\sqrt{3}-\sqrt{6}&space;\right&space;)}{-3}-\frac{4\sqrt{3}\left&space;(&space;\sqrt{6}-\sqrt{2}&space;\right&space;)}{4}+\frac{\sqrt{6}\left&space;(&space;\sqrt{2}-\sqrt{3}&space;\right&space;)}{-1}$ $\fn_cm&space;=-\sqrt{2}\left&space;(&space;\sqrt{3}-\sqrt{6}&space;\right&space;)-\sqrt{3}\left&space;(&space;\sqrt{6}-\sqrt{2}&space;\right&space;)-\sqrt{6}\left&space;(&space;\sqrt{2}-\sqrt{3}&space;\right&space;)$ $\fn_cm&space;=-\sqrt{6}+\sqrt{12}-\sqrt{18}+\sqrt{6}-\sqrt{12}+\sqrt{18}$ $\fn_cm&space;=0$ উত্তরঃ নির্ণেয়   $\fn_cm&space;{\color{DarkGreen}&space;\frac{3\sqrt{x}}{\sqrt{y}+\sqrt{z}}-\frac{4\sqrt{y}}{\sqrt{z}+\sqrt{x}}+\frac{\sqrt{z}}{\sqrt{x}+\sqrt{y}}}$  এর মান $\fn_jvn&space;{\color{DarkGreen}&space;0}$ .

Koshe dekhi 9.3 class 10

 Q4.  $\fn_cm&space;{\color{Blue}&space;x=\sqrt{7}+\sqrt{6}}$  হলে, (i) $\fn_cm&space;{\color{Blue}&space;x-\frac{1}{x}}$ (ii) $\fn_cm&space;{\color{Blue}&space;x+\frac{1}{x}}$ (iii) $\fn_cm&space;{\color{Blue}&space;{{x}^{2}}+\frac{1}{{{x}^{2}}}}$ (iv) $\fn_cm&space;{\color{Blue}&space;{{x}^{3}}+\frac{1}{{{x}^{3}}}}$ এদের সরলতম মান নির্ণয় করি। সমাধানঃ  $\fn_cm&space;\because&space;x=\left&space;(&space;\sqrt{7}&space;+\sqrt{6}\right&space;)$ $\fn_cm&space;\therefore&space;\frac{1}{x}=\frac{1}{\left&space;(&space;\sqrt{7}+\sqrt{6}&space;\right&space;)}$ $\fn_cm&space;=\frac{1}{\left&space;(&space;\sqrt{7}+\sqrt{6}&space;\right&space;)}\times&space;\frac{\left&space;(&space;\sqrt{7}-\sqrt{6}&space;\right&space;)}{\left&space;(&space;\sqrt{7}-\sqrt{6}&space;\right&space;)}$ [ হরের করণী নিরসন করে পাই ] $\fn_cm&space;=\frac{\left&space;(&space;\sqrt{7}-\sqrt{6}&space;\right&space;)}{\left&space;(&space;\sqrt{7}&space;\right&space;)^{2}-\left&space;(&space;\sqrt{6}&space;\right&space;)^{2}}$ $\fn_cm&space;=\frac{\left&space;(&space;\sqrt{7}-\sqrt{6}&space;\right&space;)}{7-6}$ $\fn_cm&space;=\left&space;(&space;\sqrt{7}-\sqrt{6}&space;\right&space;)$   সমাধান (i) : এখন, (i) $\fn_cm&space;x-\frac{1}{x}$ $\fn_jvn&space;=\left&space;(&space;\sqrt{7}+\sqrt{6}&space;\right&space;)-\left&space;(&space;\sqrt{7}-\sqrt{6}&space;\right&space;)$ [ x ও $\fn_cm&space;{\color{Blue}&space;\frac{1}{x}}$ এর মান  বসিয়ে পাই ] $\fn_cm&space;=\sqrt{7}+\sqrt{6}-\sqrt{7}+\sqrt{6}$ $\fn_cm&space;=2\sqrt{6}$ উত্তরঃ নির্ণেয় $\fn_cm&space;{\color{DarkGreen}&space;x-\frac{1}{x}}$ এর  সরলতম মান $\fn_cm&space;{\color{DarkGreen}&space;2\sqrt{6}}$   সমাধান (ii) : (ii) $\fn_cm&space;x+\frac{1}{x}$ $\fn_cm&space;=\left&space;(&space;\sqrt{7}+\sqrt{6}&space;\right&space;)+\left&space;(&space;\sqrt{7}-\sqrt{6}&space;\right&space;)$ [ x ও $\fn_cm&space;{\color{Blue}&space;\frac{1}{x}}$ এর মান  বসিয়ে পাই ] $\fn_cm&space;=\sqrt{7}+\sqrt{6}+\sqrt{7}-\sqrt{6}$ $\fn_cm&space;=2\sqrt{7}$ উত্তরঃ নির্ণেয় $\fn_cm&space;{\color{DarkGreen}&space;x+\frac{1}{x}}$ এর  সরলতম মান $\fn_cm&space;{\color{DarkGreen}&space;2\sqrt{7}}$   সমাধান (iii) : (iii) $\fn_cm&space;{{x}^{2}}+\frac{1}{{{x}^{2}}}$ $\fn_cm&space;=\left&space;(&space;x+\frac{1}{x}&space;\right&space;)^{2}-2\cdot&space;x\cdot&space;\frac{1}{x}$ $\fn_cm&space;=\left&space;(&space;2\sqrt{7}&space;\right&space;)^{2}-2$    [ $\fn_cm&space;{\color{Blue}&space;x+\frac{1}{x}}$ এর মান বসিয়ে পাই ] $\fn_cm&space;=28-2$ $\fn_cm&space;=26$ উত্তরঃ নির্ণেয় $\fn_cm&space;{\color{DarkGreen}&space;{{x}^{2}}+\frac{1}{{{x}^{2}}}}$ এর  সরলতম মান $\fn_cm&space;{\color{DarkGreen}&space;26}$     সমাধান (iv) : (iv) $\fn_cm&space;{{x}^{3}}+\frac{1}{{{x}^{3}}}$ $\fn_cm&space;=\left&space;(&space;x+\frac{1}{x}&space;\right&space;)^{3}-3\cdot&space;x\cdot&space;\frac{1}{x}\cdot&space;\left&space;(&space;x+\frac{1}{x}&space;\right&space;)$ $\fn_cm&space;=\left&space;(&space;2\sqrt{7}&space;\right&space;)^{3}-3\times&space;2\sqrt{7}$      [ $\fn_cm&space;{\color{Blue}&space;x+\frac{1}{x}}$ এর মান বসিয়ে পাই ] $\fn_cm&space;=56\sqrt{7}-6\sqrt{7}$ $\fn_cm&space;=50\sqrt{7}$ উত্তরঃ নির্ণেয় $\fn_cm&space;{\color{DarkGreen}&space;{{x}^{3}}+\frac{1}{{{x}^{3}}}}$ এর  সরলতম মান $\fn_cm&space;{\color{DarkGreen}&space;50\sqrt{7}}$.

Koshe dekhi 9.3 class 10

 Q5. সরল করি :  $\small&space;{\color{Blue}&space;\frac{x+\sqrt{{{x}^{2}}-1}}{x-\sqrt{{{x}^{2}}-1}}+\frac{x-\sqrt{{{x}^{2}}-1}}{x+\sqrt{{{x}^{2}}-1}}}$  সরলফল $\fn_cm&space;{\color{Blue}&space;14}$ হলে, $\fn_cm&space;{\color{Blue}&space;x}$ -এর মান কী কী হবে হিসাব করে লিখি। সমাধানঃ  $\fn_cm&space;\frac{x+\sqrt{{{x}^{2}}-1}}{x-\sqrt{{{x}^{2}}-1}}+\frac{x-\sqrt{{{x}^{2}}-1}}{x+\sqrt{{{x}^{2}}-1}}=14$ বা, $\fn_cm&space;\frac{\left&space;(&space;x+\sqrt{x^{2}-1}&space;\right&space;)^{2}+\left&space;(&space;x-\sqrt{x^{2}-1}&space;\right&space;)^{2}}{\left&space;(&space;x-\sqrt{x^{2}-1}&space;\right&space;)\left&space;(&space;x+\sqrt{x^{2}-1}&space;\right&space;)}=14$ বা, $\fn_cm&space;\frac{2\left&space;[\left&space;(&space;x\right&space;)^{2}&space;+\left&space;(&space;\sqrt{x^{2}-1}&space;\right&space;)&space;^{2}\right&space;]}{\left&space;(&space;x\right&space;)^{2}&space;-\left&space;(&space;\sqrt{x^{2}-1}&space;\right&space;)&space;^{2}}=14$  $\fn_cm&space;{\color{Blue}\left&space;[\because&space;\left&space;(&space;i&space;\right&space;)&space;\left&space;(&space;a+b&space;\right&space;)^{2}+\left&space;(&space;a-b&space;\right&space;)^{2}=2\left&space;(a^{2}+b&space;^{2}&space;\right&space;),&space;\left&space;(&space;ii&space;\right&space;)&space;\left&space;(&space;a-b&space;\right&space;)\left&space;(&space;a+b&space;\right&space;)=a^{2}-b^{2}\right&space;]}$ বা, $\fn_cm&space;\frac{2\left&space;(&space;x^{2}&space;+&space;x^{2}-1\right&space;)}{&space;x^{2}-&space;x^{2}+1}=14$ বা, $\fn_cm&space;\frac{2\left&space;(&space;2x^{2}-1&space;\right&space;)}{1}=14$ বা, $\fn_cm&space;2x^{2}-1=\frac{14}{2}$ বা, $\fn_cm&space;2x^{2}=7+1$ বা, $\fn_cm&space;2x^{2}=8$ বা, $\fn_cm&space;x^{2}=\frac{8}{2}=4$ বা, $\fn_cm&space;x=\pm&space;\sqrt{4}$ $\fn_cm&space;\therefore&space;x=\pm&space;2$ উত্তরঃ নির্ণেয় $\fn_cm&space;{\color{DarkGreen}&space;x}$ -এর মান $\fn_cm&space;{\color{DarkGreen}&space;\pm&space;2}$ হবে।

Koshe dekhi 9.3 class 10

 Q6. যদি   $\fn_cm&space;{\color{Blue}&space;a=\frac{\sqrt{5}+1}{\sqrt{5}-1}}$  ও   $\fn_cm&space;{\color{Blue}&space;b=\frac{\sqrt{5}-1}{\sqrt{5}+1}}$  হয়, তবে নীচের মানগুলি নির্ণয় করি। (i) $\fn_cm&space;{\color{Blue}&space;\frac{{{a}^{2}}+ab+{{b}^{2}}}{{{a}^{2}}-ab+{{b}^{2}}}}$ (ii) $\fn_cm&space;{\color{Blue}&space;\frac{{{\left(&space;a-b&space;\right)}^{3}}}{{{\left(&space;a+b&space;\right)}^{3}}}}$ (iii) $\fn_cm&space;{\color{Blue}&space;\frac{3{{a}^{2}}+5ab+3{{b}^{2}}}{3{{a}^{2}}-5ab+3{{b}^{2}}}}$ (iv) $\fn_cm&space;{\color{Blue}&space;\frac{{{a}^{3}}+{{b}^{3}}}{{{a}^{3}}-{{b}^{3}}}}$ সমাধানঃ  $\fn_cm&space;a=\frac{\sqrt{5}+1}{\sqrt{5}-1}$ ,   $\fn_cm&space;b=\frac{\sqrt{5}-1}{\sqrt{5}+1}$   ∴ a + b $\fn_cm&space;=\frac{\sqrt{5}+1}{\sqrt{5}-1}+\frac{\sqrt{5}-1}{\sqrt{5}+1}$ $\fn_cm&space;=\frac{\left&space;(&space;\sqrt{5}+1&space;\right&space;)^{2}+\left&space;(&space;\sqrt{5}-1&space;\right&space;)^{2}}{\left&space;(&space;\sqrt{5}-1&space;\right&space;)\left&space;(&space;\sqrt{5}+1&space;\right&space;)}$ $\fn_cm&space;=\frac{2\left&space;[&space;\left&space;(&space;\sqrt{5}&space;\right&space;)^{2}+\left&space;(&space;1&space;\right&space;)^{2}&space;\right&space;]}{\left&space;(&space;\sqrt{5}&space;\right&space;)^{2}-\left&space;(&space;1&space;\right&space;)^{2}}$ $\fn_cm&space;{\color{Blue}\left&space;[\because&space;\left&space;(&space;i&space;\right&space;)&space;\left&space;(&space;a+b&space;\right&space;)^{2}+\left&space;(&space;a-b&space;\right&space;)^{2}=2\left&space;(a^{2}+b&space;^{2}&space;\right&space;),&space;\left&space;(&space;ii&space;\right&space;)&space;\left&space;(&space;a-b&space;\right&space;)\left&space;(&space;a+b&space;\right&space;)=a^{2}-b^{2}\right&space;]}$ $\fn_cm&space;=\frac{2\left&space;(&space;5+1&space;\right&space;)}{\left&space;(&space;5-1&space;\right&space;)}$ $\fn_cm&space;=\frac{2\times&space;6}{4}$ $\fn_cm&space;=3$ এবং, $\fn_cm&space;\left&space;(&space;a-b&space;\right&space;)$ $\fn_cm&space;=\frac{\sqrt{5}+1}{\sqrt{5}-1}-\frac{\sqrt{5}-1}{\sqrt{5}+1}$ $\fn_cm&space;=\frac{\left&space;(&space;\sqrt{5}+1&space;\right&space;)^{2}-\left&space;(&space;\sqrt{5}-1&space;\right&space;)^{2}}{\left&space;(&space;\sqrt{5}-1&space;\right&space;)\left&space;(&space;\sqrt{5}+1&space;\right&space;)}$ $\fn_cm&space;=\frac{4\times&space;\sqrt{5}\times&space;1}{\left&space;(&space;\sqrt{5}&space;\right&space;)^{2}-\left&space;(&space;1&space;\right&space;)^{2}}$ $\fn_cm&space;{\color{Blue}\left&space;[\because&space;\left&space;(&space;i&space;\right&space;)&space;\left&space;(&space;a+b&space;\right&space;)^{2}-\left&space;(&space;a-b&space;\right&space;)^{2}=4ab,&space;\left&space;(&space;ii&space;\right&space;)&space;\left&space;(&space;a-b&space;\right&space;)\left&space;(&space;a+b&space;\right&space;)=a^{2}-b^{2}\right&space;]}$ $\fn_cm&space;=\frac{4\sqrt{5}}{\left&space;(&space;5-1&space;\right&space;)}$ $\fn_cm&space;=\frac{4\sqrt{5}}{4}$ $\fn_cm&space;=\sqrt{5}$ আবার, $\fn_cm&space;ab=\frac{\left&space;(\sqrt{5}+1&space;\right&space;)}{\left&space;(\sqrt{5}-1&space;\right&space;)}\times&space;\frac{\left&space;(\sqrt{5}-1&space;\right&space;)}{\left&space;(\sqrt{5}+1&space;\right&space;)}$ $\fn_cm&space;\therefore&space;ab=1$   সমাধান (i) : এখন, (i) $\fn_cm&space;\frac{{{a}^{2}}+ab+{{b}^{2}}}{{{a}^{2}}-ab+{{b}^{2}}}$ $\fn_cm&space;=\frac{\left&space;\{&space;\left&space;(&space;a+b&space;\right&space;)^{2}-2ab&space;\right&space;\}+ab}{\left&space;\{&space;\left&space;(&space;a+b&space;\right&space;)^{2}-2ab&space;\right&space;\}-ab}&space;{\color{Blue}&space;\left&space;[&space;\because&space;a^{2}+b^{2}=\left&space;(&space;a+b&space;\right&space;)^{2}-2ab&space;\right&space;]}$ $\fn_cm&space;=\frac{\left&space;(&space;3&space;\right&space;)^{2}-2\times&space;1+1}{\left&space;(&space;3&space;\right&space;)^{2}-2\times&space;1-1}$ [ $\fn_cm&space;{\color{Blue}&space;a+b=3}$ এবং $\fn_cm&space;{\color{Blue}&space;ab=1}$ বসিয়ে পাই ] $\fn_cm&space;=\frac{9-2+1}{9-2-1}$ $\fn_cm&space;=\frac{8}{6}$ $\fn_cm&space;=\frac{4}{3}$ $\fn_cm&space;=1\frac{1}{3}$ উত্তরঃ নির্ণেয়  $\fn_cm&space;{\color{DarkGreen}&space;\frac{{{a}^{2}}+ab+{{b}^{2}}}{{{a}^{2}}-ab+{{b}^{2}}}}$ এর মান $\fn_cm&space;{\color{DarkGreen}&space;1\frac{1}{3}}$     সমাধান (ii) : (ii) $\fn_cm&space;\frac{{{\left(&space;a-b&space;\right)}^{3}}}{{{\left(&space;a+b&space;\right)}^{3}}}$ $\fn_cm&space;=\frac{\left&space;(&space;\sqrt{5}&space;\right&space;)^{3}}{\left&space;(&space;3&space;\right&space;)^{3}}$   $\fn_cm&space;{\color{Blue}[\because&space;a-b=\sqrt{5}}$ এবং $\fn_cm&space;{\color{Blue}&space;a+b=3]}$  $\fn_cm&space;=\frac{5\sqrt{5}}{27}$ উত্তরঃ নির্ণেয় $\fn_cm&space;{\color{DarkGreen}&space;\frac{{{\left(&space;a-b&space;\right)}^{3}}}{{{\left(&space;a+b&space;\right)}^{3}}}}$  এর মান $\fn_cm&space;{\color{DarkGreen}&space;\frac{5\sqrt{5}}{27}}$   সমাধান (iii) : (iii) $\fn_cm&space;\frac{3{{a}^{2}}+5ab+3{{b}^{2}}}{3{{a}^{2}}-5ab+3{{b}^{2}}}$ $\fn_cm&space;=\frac{3\left&space;(&space;a^{2}&space;+b^{2}&space;\right&space;)+5ab}{3\left&space;(&space;a^{2}&space;+b^{2}&space;\right&space;)-5ab}$ $\fn_cm&space;=\frac{3\left&space;[\left&space;(&space;a+b&space;\right&space;)^{2}-2ab&space;\right&space;]+5ab}{3\left&space;[\left&space;(&space;a+b&space;\right&space;)^{2}-2ab&space;\right&space;]-5ab}&space;{\color{Blue}&space;\left&space;[&space;\because&space;a^{2}+b^{2}=\left&space;(&space;a+b&space;\right&space;)^{2}-2ab&space;\right&space;]}$ $\fn_cm&space;=\frac{3\left&space;[\left&space;(&space;3&space;\right&space;)^{2}-2\times&space;1&space;\right&space;]+1}{3\left&space;[\left&space;(&space;3&space;\right&space;)^{2}-2\times&space;1&space;\right&space;]-1}$ [ $\fn_cm&space;{\color{Blue}&space;a+b=3}$ এবং $\fn_cm&space;{\color{Blue}&space;ab=1}$ বসিয়ে পাই ] $\fn_cm&space;=\frac{3\left&space;[&space;9-2&space;\right&space;]+5\times&space;1}{3\left&space;[&space;9-2&space;\right&space;]-5\times&space;1}$ $\fn_cm&space;=\frac{3\times&space;7+5}{3\times&space;7-5}$ $\fn_cm&space;=\frac{26}{16}$ $\fn_cm&space;=\frac{13}{8}$ $\fn_cm&space;=1\frac{5}{8}$ উত্তরঃ নির্ণেয় $\fn_cm&space;{\color{DarkGreen}&space;\frac{3{{a}^{2}}+5ab+3{{b}^{2}}}{3{{a}^{2}}-5ab+3{{b}^{2}}}}$  এর মান $\fn_cm&space;{\color{DarkGreen}&space;1\frac{5}{8}}$     সমাধান (iv) : (iv) $\fn_cm&space;\frac{{{a}^{3}}+{{b}^{3}}}{{{a}^{3}}-{{b}^{3}}}$ $\fn_cm&space;=\frac{\left&space;(&space;a+b&space;\right&space;)^{3}-3ab\left&space;(&space;a+b&space;\right&space;)}{\left&space;(&space;a-b&space;\right&space;)^{3}+3ab\left&space;(&space;a-b&space;\right&space;)}$ $\fn_cm&space;=\frac{\left&space;(&space;3&space;\right&space;)^{3}-3\times&space;1\times&space;3}{\left&space;(&space;\sqrt{5}&space;\right&space;)^{3}+3\times&space;1\times&space;\sqrt{5}}$ $\fn_cm&space;=\frac{27-9}{5\sqrt{5}+3\sqrt{5}}$ $\fn_cm&space;=\frac{18}{8\sqrt{5}}$ $\fn_cm&space;=\frac{9}{4\sqrt{5}}$ $\fn_cm&space;=\frac{9\times&space;\sqrt{5}}{4\sqrt{5}\times&space;\sqrt{5}}$ $\fn_cm&space;=\frac{9\sqrt{5}}{20}$ উত্তরঃ নির্ণেয়  $\fn_cm&space;{\color{DarkGreen}&space;\frac{{{a}^{3}}+{{b}^{3}}}{{{a}^{3}}-{{b}^{3}}}}$  এর মান $\fn_cm&space;{\color{DarkGreen}&space;\frac{9\sqrt{5}}{20}}$

Koshe dekhi 9.3 class 10

 Q7. যদি  $\fn_cm&space;{\color{Blue}&space;x=2+\sqrt{3}}$  এবং  $\fn_cm&space;{\color{Blue}&space;y=2-\sqrt{3}}$  হয়, তবে নিম্নলিখিতগুলির সরলতম মান নির্ণয় করি : (a) (i) $\fn_cm&space;{\color{Blue}&space;x-\frac{1}{x}}$ (ii) $\fn_cm&space;{\color{Blue}&space;{{y}^{2}}+\frac{1}{{{y}^{2}}}}$ (iii) $\fn_cm&space;{\color{Blue}&space;{{x}^{3}}-\frac{1}{{{x}^{3}}}}$ (iv) $\fn_cm&space;{\color{Blue}&space;xy+\frac{1}{xy}}$ (b) $\fn_cm&space;{\color{Blue}&space;3{{x}^{2}}-5xy+3{{y}^{2}}}$ সমাধানঃ  $\fn_cm&space;\because&space;x=2+\sqrt{3}$ $\fn_cm&space;\therefore&space;\frac{1}{x}=\frac{1}{2+\sqrt{3}}$ বা, $\fn_cm&space;\frac{1}{x}=\frac{\left&space;(2-\sqrt{3}&space;\right&space;)}{\left&space;(2+\sqrt{3}&space;\right&space;)\left&space;(2-\sqrt{3}&space;\right&space;)}$ [ হরের করণী নিরসন করে পাই ] বা, $\fn_cm&space;\frac{1}{x}=\frac{2-\sqrt{3}}{\left&space;(&space;2&space;\right&space;)&space;^{2}-\left&space;(&space;\sqrt{3}&space;\right&space;)^{2}}$ $\fn_cm&space;{\color{Blue}&space;\left&space;[&space;\because&space;\left&space;(&space;a+b&space;\right&space;)\left&space;(&space;a-b&space;\right&space;)=a^{2}-b^{2}&space;\right&space;]}$ বা, $\fn_cm&space;\frac{1}{x}=\frac{2-\sqrt{3}}{4-3}\Rightarrow&space;2-\sqrt{3}$ $\fn_cm&space;\therefore&space;\frac{1}{x}=2-\sqrt{3}$ একইভাবে, $\fn_cm&space;\because&space;y=2-\sqrt{3}$ $\fn_cm&space;\therefore&space;\frac{1}{y}=\frac{1}{2-\sqrt{3}}$ বা, $\fn_cm&space;\frac{1}{y}=\frac{\left&space;(2+\sqrt{3}&space;\right&space;)}{\left&space;(2+\sqrt{3}&space;\right&space;)\left&space;(2-\sqrt{3}&space;\right&space;)}$ [ হরের করণী নিরসন করে পাই ] বা, $\fn_cm&space;\frac{1}{y}=\frac{2+\sqrt{3}}{\left&space;(&space;2&space;\right&space;)&space;^{2}-\left&space;(&space;\sqrt{3}&space;\right&space;)^{2}}$ $\fn_cm&space;{\color{Blue}&space;\left&space;[&space;\because&space;\left&space;(&space;a+b&space;\right&space;)\left&space;(&space;a-b&space;\right&space;)=a^{2}-b^{2}&space;\right&space;]}$ বা, $\fn_cm&space;\frac{1}{y}=\frac{2+\sqrt{3}}{4-3}\Rightarrow&space;2+\sqrt{3}$ $\fn_cm&space;\therefore&space;\frac{1}{y}=2+\sqrt{3}$   সমাধান (i) : এখন, (i) $\fn_cm&space;{\color{Blue}&space;x-\frac{1}{x}}$ $\fn_cm&space;=\left&space;(&space;2+\sqrt{3}&space;\right&space;)-\left&space;(&space;2-\sqrt{3}&space;\right&space;)$ [ $\fn_cm&space;{\color{Blue}&space;x}$ ও $\fn_cm&space;{\color{Blue}&space;\frac{1}{x}}$ -এর মান বসিয়ে পাই ] $\fn_cm&space;=2+\sqrt{3}-2+\sqrt{3}$ $\fn_cm&space;=2\sqrt{3}$ উত্তরঃ নির্ণেয় $\fn_cm&space;{\color{DarkGreen}&space;x-\frac{1}{x}}$ এর  সরলতম মান $\fn_cm&space;{\color{DarkGreen}&space;2\sqrt{3}}$ .   সমাধান (ii) : (ii) $\fn_cm&space;{\color{Blue}&space;{{y}^{2}}+\frac{1}{{{y}^{2}}}}$ $\fn_cm&space;\because&space;y=2-\sqrt{3}$ এবং $\fn_cm&space;\frac{1}{y}=2+\sqrt{3}$ $\fn_cm&space;\therefore&space;y+\frac{1}{y}$ $\fn_cm&space;=\left&space;(&space;2-\sqrt{3}&space;\right&space;)+\left&space;(&space;2+\sqrt{3}&space;\right&space;)$ [ $\fn_cm&space;{\color{Blue}&space;y}$ ও $\fn_cm&space;{\color{Blue}&space;\frac{1}{y}}$ -এর মান বসিয়ে পাই ] $\fn_cm&space;=2-\sqrt{3}+2+\sqrt{3}$ $\fn_cm&space;=4$ সুতরাং, $\fn_cm&space;y+\frac{1}{y}=4$ এখন,  $\fn_cm&space;{{y}^{2}}+\frac{1}{{{y}^{2}}}$ $\fn_cm&space;=\left&space;(&space;y+\frac{1}{y}&space;\right&space;)^{2}-2\cdot&space;y\cdot&space;\frac{1}{y}$ $\fn_cm&space;=4^{2}-2\;&space;{\color{Blue}&space;\left&space;[\because&space;y+\frac{1}{y}=4&space;\right&space;]}$ $\fn_cm&space;=16-2\Rightarrow&space;14$ উত্তরঃ নির্ণেয় $\fn_cm&space;{\color{DarkGreen}&space;{{y}^{2}}+\frac{1}{{{y}^{2}}}}$ এর  সরলতম মান $\fn_cm&space;{\color{DarkGreen}&space;14}$.   সমাধান (iii) : (iii) $\fn_cm&space;{\color{Blue}&space;{{x}^{3}}-\frac{1}{{{x}^{3}}}}$ $\fn_cm&space;=\left&space;(&space;x-\frac{1}{x}&space;\right&space;)^{3}+3\cdot&space;x\cdot&space;\frac{1}{x}\left&space;(x-\frac{1}{x}&space;\right&space;)\;&space;{\color{Blue}&space;\left&space;[&space;\because&space;x-\frac{1}{x}=2\sqrt{3}&space;\right&space;]}$ $\fn_cm&space;=\left&space;(&space;2\sqrt{3}&space;\right&space;)^{3}+3\times&space;2\sqrt{3}$ $\fn_cm&space;=24\sqrt{3}+6\sqrt{3}$ $\fn_cm&space;=30\sqrt{3}$ উত্তরঃ নির্ণেয় $\fn_cm&space;{\color{DarkGreen}&space;{{x}^{3}}-\frac{1}{{{x}^{3}}}}$ এর  সরলতম মান $\fn_cm&space;{\color{DarkGreen}&space;30\sqrt{3}}$.   সমাধান (iv) : (iv) $\fn_cm&space;{\color{Blue}&space;xy+\frac{1}{xy}}$ $\fn_cm&space;\because&space;x=2+\sqrt{3}$ এবং  $\fn_cm&space;y=2-\sqrt{3}$  $\fn_cm&space;\therefore&space;xy=\left&space;(&space;2+\sqrt{3}&space;\right&space;)\left&space;(&space;2-\sqrt{3}&space;\right&space;)$ $\fn_cm&space;=\left&space;(&space;2&space;\right&space;)^{2}-\left&space;(&space;2\sqrt{3}&space;\right&space;)^{2}$ $\fn_cm&space;=4-3$ $\fn_cm&space;=1$ $\fn_cm&space;\therefore&space;xy+\frac{1}{xy}$ $\fn_cm&space;=1+\frac{1}{1}$ [ $\fn_cm&space;{\color{Blue}&space;xy=1}$ বসিয়ে পাই ] $\fn_cm&space;=1+1$ $\fn_cm&space;=2$ উত্তরঃ নির্ণেয় $\fn_cm&space;{\color{DarkGreen}&space;xy+\frac{1}{xy}}$ এর  সরলতম মান 2.   সমাধান (b) : (b) $\fn_cm&space;{\color{Blue}&space;3{{x}^{2}}-5xy+3{{y}^{2}}}$ $\fn_cm&space;\because&space;x=2+\sqrt{3}$ এবং  $\fn_cm&space;y=2-\sqrt{3}$  $\fn_cm&space;\therefore&space;x+y=\left&space;(&space;2+\sqrt{3}&space;\right&space;)+\left&space;(&space;2-\sqrt{3}&space;\right&space;)$ $\fn_cm&space;=2+\sqrt{3}+2-\sqrt{3}$ $\fn_cm&space;=4$ এখন, $\fn_cm&space;3x^{2}-5xy+3y^{2}$ $\fn_cm&space;=3\left&space;(&space;x^{2}+y^{2}&space;\right&space;)-5xy$ $\fn_cm&space;=3\left&space;\{&space;\left&space;(&space;x+y&space;\right&space;)^{2}-2xy&space;\right&space;\}-5xy$ $\fn_cm&space;=3\left&space;\{&space;\left&space;(&space;4&space;\right&space;)^{2}-2\times&space;1&space;\right&space;\}-5\times&space;1$ $\fn_cm&space;{\color{Blue}&space;\left&space;[&space;\because&space;x+y=4,\;&space;xy=1&space;\right&space;]}$ $\fn_cm&space;=3\left&space;(&space;16-2&space;\right&space;)-5$ $\fn_cm&space;=3\times&space;14-5\Rightarrow&space;42-5$ $\fn_cm&space;=37$ উত্তরঃ নির্ণেয় $\fn_cm&space;{\color{DarkGreen}&space;3{{x}^{2}}-5xy+3{{y}^{2}}}$ এর  সরলতম মান $\fn_cm&space;{\color{DarkGreen}37}.$

Koshe dekhi 9.3 class 10

 Q8.  $\small&space;{\color{Blue}&space;x=\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}}}$  এবং  $\small&space;{\color{Blue}&space;xy=1}$  হলে, দেখাই যে, $\small&space;{\color{Blue}&space;\frac{{{x}^{2}}+xy+{{y}^{2}}}{{{x}^{2}}-xy+{{y}^{2}}}=\frac{12}{11}}$ সমাধানঃ  প্রদত্ত, $\fn_cm&space;\because&space;x=\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}}$ এবং  $\fn_cm&space;xy=1$  $\fn_cm&space;\therefore&space;y=\frac{1}{x}$ $\fn_cm&space;=\frac{1}{\left&space;(\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}}&space;\right&space;)}$ $\fn_cm&space;=\frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}}$ এবং, $\fn_cm&space;x+y$ $\fn_cm&space;=\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}}+\frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}}$ [ x  ও  y  এর মান বসিয়ে পাই ] $\fn_cm&space;=\frac{\left&space;(&space;\sqrt{7}+\sqrt{3}&space;\right&space;)^{2}+\left&space;(&space;\sqrt{7}-\sqrt{3}&space;\right&space;)^{2}}{\left&space;(&space;\sqrt{7}+\sqrt{3}&space;\right&space;)\left&space;(&space;\sqrt{7}-\sqrt{3}&space;\right&space;)}$ $\fn_cm&space;=\frac{2\left&space;[&space;\left&space;(&space;\sqrt{7}&space;\right&space;)^{2}+\left&space;(&space;\sqrt{3}&space;\right&space;)^{2}&space;\right&space;]}{\left&space;(&space;\sqrt{7}&space;\right&space;)^{2}-\left&space;(&space;\sqrt{3}&space;\right&space;)^{2}}$ $\fn_cm&space;{\color{Blue}\left&space;[\because&space;\left&space;(&space;i&space;\right&space;)&space;\left&space;(&space;a+b&space;\right&space;)^{2}+\left&space;(&space;a-b&space;\right&space;)^{2}=2\left&space;(a^{2}+b&space;^{2}&space;\right&space;),&space;\left&space;(&space;ii&space;\right&space;)&space;\left&space;(&space;a-b&space;\right&space;)\left&space;(&space;a+b&space;\right&space;)=a^{2}-b^{2}\right&space;]}$ $\fn_cm&space;=\frac{2\left&space;(&space;7+3&space;\right&space;)}{\left&space;(&space;7-3&space;\right&space;)}$ $\fn_cm&space;=\frac{20}{4}=5$ $\fn_cm&space;\therefore&space;x+y=5$ বামপক্ষ: $\fn_cm&space;\frac{{{x}^{2}}+xy+{{y}^{2}}}{{{x}^{2}}-xy+{{y}^{2}}}$ $\fn_cm&space;=\frac{\left&space;(&space;x^{2}+y^{2}&space;\right&space;)+xy}{\left&space;(&space;x^{2}-y^{2}&space;\right&space;)-xy}$ $\fn_cm&space;=\frac{\left&space;[&space;\left&space;(&space;x+y&space;\right&space;)^{2}-2xy&space;\right&space;]+xy}{\left&space;[&space;\left&space;(&space;x+y&space;\right&space;)^{2}-2xy&space;\right&space;]-xy}$ $\fn_cm&space;=\frac{\left&space;(&space;5&space;\right&space;)^{2}-2\times&space;1+1}{\left&space;(&space;5&space;\right&space;)^{2}-2\times&space;1-1}$ $\fn_cm&space;{\color{Blue}&space;\left&space;[&space;\because&space;x+y=5,xy=1&space;\right&space;]}$ $\fn_cm&space;=\frac{25-2+1}{25-2-1}$ $\fn_cm&space;=\frac{24}{22}$ $\fn_cm&space;=\frac{12}{11}=$ ডানপক্ষ ( প্রমাণিত )

Koshe dekhi 9.3 class 10

 Q9.  $\fn_cm&space;{\color{Blue}&space;\left(&space;\sqrt{7}+1&space;\right)}$  এবং  $\fn_cm&space;{\color{Blue}&space;\left(&space;\sqrt{5}+\sqrt{3}&space;\right)}$  এর মধ্যে কোনটি বড়ো লিখি। সমাধানঃ  $\fn_jvn&space;\left(&space;\sqrt{7}+1&space;\right)^{2}$ [ বর্গ করে পাই ] $\fn_jvn&space;=7+2\sqrt{7}+1$ $\fn_jvn&space;=8+2\sqrt{7}$ আবার,   $\fn_jvn&space;\left(&space;\sqrt{5}+\sqrt{3}&space;\right)^{2}$  [ বর্গ করে পাই ] $\fn_jvn&space;=5+2\sqrt{15}+3$ $\fn_jvn&space;=8+2\sqrt{15}$ $\fn_jvn&space;\because&space;\sqrt{15}>\sqrt{7}$ $\fn_jvn&space;\therefore&space;\left&space;(&space;\sqrt{5}+\sqrt{3}&space;\right&space;)^{2}>\left(&space;\sqrt{7}+1&space;\right)^{2}$ বা, $\fn_jvn&space;\left&space;(&space;\sqrt{5}+\sqrt{3}&space;\right&space;)>\left(&space;\sqrt{7}+1&space;\right)$ উত্তরঃ নির্ণেয় $\fn_jvn&space;{\color{DarkGreen}&space;\left(&space;\sqrt{7}+1&space;\right)}$  এবং  $\fn_jvn&space;{\color{DarkGreen}&space;\left(&space;\sqrt{5}+\sqrt{3}&space;\right)}$  এর মধ্যে $\fn_jvn&space;{\color{DarkGreen}&space;\left(&space;\sqrt{5}+\sqrt{3}&space;\right)}$  বড়ো।

Koshe dekhi 9.3 class 10

 Q10. অতিরিক্ত সংক্ষিপ্ত উত্তরধর্মী প্রশ্ন (V.S.A.) (A) বহুবিকল্পীয় প্রশ্ন (M.C.Q.) (i)  $\small&space;{\color{Blue}&space;x=2+\sqrt{3}}$   হলে,  $\small&space;{\color{Blue}&space;x+\frac{1}{x}}$  এর মান – (a) 2 (b) 2√3 (c) 4 (d) 2 − √3 সমাধানঃ  $\fn_jvn&space;x=2+\sqrt{3}$ , $\fn_jvn&space;\frac{1}{x}=\frac{1}{2+\sqrt{3}}$ বা, $\fn_jvn&space;\frac{1}{x}=\frac{\left&space;(&space;(2-\sqrt{3}&space;\right&space;)}{\left&space;(2+\sqrt{3}&space;\right&space;)\left&space;(2-\sqrt{3}&space;\right&space;)}$ [ হরের করণী নিরসন করে পাই ] বা, $\fn_jvn&space;\frac{1}{x}=\frac{2-\sqrt{3}}{\left&space;[&space;\left&space;(&space;2&space;\right&space;)&space;^{2}-\left&space;(&space;\sqrt{3}&space;\right&space;)^{2}\right&space;]}$ $\fn_jvn&space;{\color{Blue}&space;\left&space;[&space;\because&space;\left&space;(&space;a+b&space;\right&space;)\left&space;(&space;a-b&space;\right&space;)=a^{2}-b^{2}&space;\right&space;]}$ $\fn_jvn&space;\therefore&space;\frac{1}{x}=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}$ $\fn_jvn&space;\therefore&space;x+\frac{1}{x}$ $\fn_jvn&space;=2+\sqrt{3}+2-\sqrt{3}$ $\fn_jvn&space;=4$ উত্তরঃ $\fn_jvn&space;{\color{DarkGreen}&space;\left&space;(&space;c&space;\right&space;)4}$

Koshe dekhi 9.3 class 10

 Q10. অতিরিক্ত সংক্ষিপ্ত উত্তরধর্মী প্রশ্ন (V.S.A.) (A) বহুবিকল্পীয় প্রশ্ন (M.C.Q.) (ii) যদি  $\small&space;{\color{Blue}&space;p+q=\sqrt{13}}$  এবং  $\small&space;{\color{Blue}&space;p-q=\sqrt{5}}$  হয়, তাহলে $\small&space;{\color{Blue}&space;pq}$ এর মান – (a) 2 (b) 18 (c) 9 (d) 8 সমাধানঃ    $\fn_jvn&space;p+q=\sqrt{13}$ বা, $\fn_jvn&space;\left&space;(&space;p+q&space;\right&space;)^{2}=\left&space;(&space;\sqrt{13}&space;\right&space;)^{2}$ [ উভয়পক্ষে বর্গ করে পাই ] বা, $\fn_jvn&space;\left&space;(&space;p+q&space;\right&space;)^{2}=13......\left&space;(&space;i&space;\right&space;)$ আবার,   $\fn_jvn&space;p-q=\sqrt{5}$  বা, $\fn_jvn&space;\left&space;(&space;p-q&space;\right&space;)^{2}=\left&space;(&space;\sqrt{5}&space;\right&space;)^{2}$ [ উভয়পক্ষে বর্গ করে পাই ] বা, $\fn_jvn&space;\left&space;(&space;p-q&space;\right&space;)^{2}=5......\left&space;(&space;ii&space;\right&space;)$ $\fn_jvn&space;\left&space;(&space;i&space;\right&space;)$ নং সমীকরণ থেকে $\fn_jvn&space;\left&space;(&space;ii&space;\right&space;)$ নং সমীকরণ বিয়োগ করে পাই  $\fn_jvn&space;\left&space;(&space;p+q&space;\right&space;)^{2}-\left&space;(&space;p-q&space;\right&space;)^{2}=13-5$ বা, $\fn_jvn&space;4pq=8&space;{\color{Blue}&space;\left&space;[&space;\because&space;\left&space;(&space;a+b&space;\right&space;)^{2}-\left&space;(&space;a-b&space;\right&space;)^{2}=4ab&space;\right&space;]}$ $\fn_jvn&space;\therefore&space;pq=\frac{8}{4}=2$ উত্তরঃ $\fn_jvn&space;{\color{DarkGreen}&space;\left&space;(&space;a&space;\right&space;)2}$

Koshe dekhi 9.3 class 10

 Q10. অতিরিক্ত সংক্ষিপ্ত উত্তরধর্মী প্রশ্ন (V.S.A.) (A) বহুবিকল্পীয় প্রশ্ন (M.C.Q.) (iii) যদি   $\small&space;{\color{Blue}&space;a+b=\sqrt{5}}$  এবং   $\small&space;{\color{Blue}&space;a-b=\sqrt{3}}$  তাহলে   $\small&space;{\color{Blue}&space;\left(&space;{{a}^{2}}+{{b}^{2}}&space;\right)}$  এর মান – (a) 8 (b) 4 (c) 2 (d) 1 সমাধানঃ  $\fn_jvn&space;a+b=\sqrt{5}$  বা, $\fn_jvn&space;\left&space;(&space;a+b&space;\right&space;)^{2}=\left&space;(&space;\sqrt{5}&space;\right&space;)^{2}$ [ উভয়পক্ষে বর্গ করে পাই ] বা, $\fn_jvn&space;\left&space;(&space;a+b&space;\right&space;)^{2}=5......\left&space;(&space;i&space;\right&space;)$ আবার, $\fn_jvn&space;a-b=\sqrt{3}$  বা, $\fn_jvn&space;\left&space;(&space;a-b&space;\right&space;)^{2}=\left&space;(&space;\sqrt{3}&space;\right&space;)^{2}$ [ উভয়পক্ষে বর্গ করে পাই ] বা, $\fn_jvn&space;\left&space;(&space;a-b&space;\right&space;)^{2}=3......\left&space;(&space;ii&space;\right&space;)$ $\fn_jvn&space;\left&space;(&space;i&space;\right&space;)$ নং সমীকরণ ও $\fn_jvn&space;\left&space;(&space;ii&space;\right&space;)$ নং সমীকরণ যোগ করে পাই  $\fn_jvn&space;\left&space;(&space;a+b&space;\right&space;)^{2}+\left&space;(&space;a-b&space;\right&space;)^{2}=5+3$ বা, $\fn_jvn&space;a^{2}+2ab+b^{2}+a^{2}-2ab+b^{2}=8$ বা, $\fn_jvn&space;2\left(&space;{{a}^{2}}+{{b}^{2}}&space;\right)=8$ $\fn_jvn&space;\therefore&space;\left(&space;{{a}^{2}}+{{b}^{2}}&space;\right)=\frac{8}{2}=4$   উত্তরঃ $\fn_jvn&space;{\color{DarkGreen}&space;\left&space;(&space;b&space;\right&space;)4}$

Koshe dekhi 9.3 class 10

 Q10. অতিরিক্ত সংক্ষিপ্ত উত্তরধর্মী প্রশ্ন (V.S.A.) (A) বহুবিকল্পীয় প্রশ্ন (M.C.Q.) (iv)  √125  থেকে  √5  বিয়োগ করলে বিয়োগফল হবে – (a) √80 (b) √120 (c) √100 (d) কোনটিই নয় সমাধানঃ  $\fn_jvn&space;\sqrt{125}$  থেকে  $\fn_jvn&space;\sqrt{5}$  বিয়োগ করলে বিয়োগফল হবে- $\fn_jvn&space;\sqrt{125}-\sqrt{5}$ $\fn_jvn&space;=\sqrt{5\times&space;5\times&space;5}-\sqrt{5}$ $\fn_jvn&space;=5\sqrt{5}-\sqrt{5}$ $\fn_jvn&space;=4\sqrt{5}$ $\fn_jvn&space;=\sqrt{4\times&space;4\times&space;5}$ $\fn_jvn&space;=\sqrt{80}$ উত্তরঃ $\fn_jvn&space;{\color{DarkGreen}&space;\left&space;(&space;a&space;\right&space;)\sqrt{80}}$

Koshe dekhi 9.3 class 10

 Q10. অতিরিক্ত সংক্ষিপ্ত উত্তরধর্মী প্রশ্ন (V.S.A.) (A) বহুবিকল্পীয় প্রশ্ন (M.C.Q.) (v)  $\small&space;{\color{Blue}&space;\left(&space;5-\sqrt{3}&space;\right)\left(&space;\sqrt{3}-1&space;\right)\left(&space;5+\sqrt{3}&space;\right)\left(&space;\sqrt{3}+1&space;\right)}$  এর গুণফল – (a) 22 (b) 44 (c) 2 (d) 11 সমাধানঃ  $\fn_jvn&space;\left(&space;5-\sqrt{3}&space;\right)\left(&space;\sqrt{3}-1&space;\right)\left(&space;5+\sqrt{3}&space;\right)\left(&space;\sqrt{3}+1&space;\right)$ এর গুণফল – $\fn_jvn&space;\left(&space;5-\sqrt{3}&space;\right)\times&space;\left(&space;\sqrt{3}-1&space;\right)\times&space;\left(&space;5+\sqrt{3}&space;\right)\times&space;\left(&space;\sqrt{3}+1&space;\right)$ $\fn_jvn&space;=\left&space;[\left(&space;5-\sqrt{3}&space;\right)\times&space;\left(&space;5+\sqrt{3}&space;\right)&space;\right&space;]\times\left&space;[&space;\left(&space;\sqrt{3}-1&space;\right)\times&space;\left(&space;\sqrt{3}+1&space;\right)&space;\right&space;]$ $\fn_jvn&space;=\left&space;[\left&space;(5&space;\right&space;)^{2}-\left&space;(\sqrt{3}&space;\right&space;)^{2}&space;\right&space;]\times\left&space;[&space;\left(&space;\sqrt{3}\right)^{2}-\left&space;(&space;1&space;\right&space;)^{2}&space;\right&space;]&space;{\color{Blue}&space;\left&space;[&space;\because&space;\left&space;(&space;a-b&space;\right&space;)\left&space;(&space;a+b&space;\right&space;)=a^{2}-b^{2}&space;\right&space;]}$ $\fn_jvn&space;=\left&space;[&space;25-3&space;\right&space;]\times&space;\left&space;[&space;3-1&space;\right&space;]$ $\fn_jvn&space;=22\times&space;2$ $\fn_jvn&space;=44$ উত্তরঃ $\fn_jvn&space;{\color{DarkGreen}&space;\left&space;(&space;b&space;\right&space;)44}$

Koshe dekhi 9.3 class 10

 (B) নীচের বিবৃতিগুলি সত্য না মিথ্যা লিখি : (i)   $\fn_jvn&space;\sqrt{75}$  এবং  $\fn_jvn&space;\sqrt{147}$  সদৃশ করণী। সমাধানঃ    $\fn_jvn&space;\sqrt{75}$  $\fn_jvn&space;=\sqrt{5\times&space;5\times&space;3}$ $\fn_jvn&space;=5\sqrt{3}$ এবং  $\fn_jvn&space;\sqrt{147}$ $\fn_jvn&space;=\sqrt{7\times&space;7\times&space;3}$ $\fn_jvn&space;=7\sqrt{3}$ ∴  $\fn_jvn&space;\sqrt{75}$  এবং  $\fn_jvn&space;\sqrt{147}$  সদৃশ করণী। উত্তরঃ বিবৃতিটি সত্য।    (ii)  $\fn_jvn&space;\sqrt{\pi}$  একটি দ্বিঘাত করণী। সমাধানঃ  বিবৃতিটি মিথ্যা।  কারণ : যে সমস্ত সংখ্যাকে  $\fn_jvn&space;\pm&space;\sqrt{a}$ আকারে লেখা যায়, যেখানে  $\fn_jvn&space;a$  হলো ধনাত্মক পূর্ণসংখ্যা।  যেহেতু, $\fn_jvn&space;\pi$ পূর্ণসংখ্যা নয় ($\fn_jvn&space;\pi$ হলো অমূলদ সংখ্যা ), তাই $\fn_jvn&space;\sqrt{\pi}$  দ্বিঘাত করোনি নয়।  উত্তরঃ বিবৃতিটি মিথ্যা।

 (C) শূন্যস্থান পূরণ করি : (i)  $\fn_jvn&space;5\sqrt{11}$  একটি _______ সংখ্যা। (মূলদ / অমূলদ) উত্তরঃ অমূলদ [ যেহেতু $\fn_jvn&space;{\color{Blue}&space;5\sqrt{11}}$ সংখ্যাটির সুনির্দিষ্ট দশমিক মান সম্পূর্ণভাবে নির্ণয় করা যায় না, তাই এটি একটি অমূলদ সংখ্যা ]   (ii)  $\fn_jvn&space;\left&space;(&space;\sqrt{3}-5&space;\right&space;)$  এর অনুবন্ধী করণী _______ । সমাধানঃ  $\fn_jvn&space;\left&space;(&space;\sqrt{3}-5&space;\right&space;)$  ও $\fn_jvn&space;\left&space;(-\sqrt{3}-5&space;\right&space;)$ এর যোগফল     $\fn_jvn&space;\left&space;(&space;\sqrt{3}-5&space;\right&space;)+\left&space;(-&space;\sqrt{3}-5&space;\right&space;)$ $\fn_jvn&space;=-10$ ( মূলদ সংখ্যা )   $\fn_jvn&space;\left&space;(&space;\sqrt{3}-5&space;\right&space;)$  ও $\fn_jvn&space;\left&space;(-\sqrt{3}-5&space;\right&space;)$ এর গুনফল  $\fn_jvn&space;\left&space;(&space;\sqrt{3}-5&space;\right&space;)\times&space;\left&space;(-&space;\sqrt{3}-5&space;\right&space;)$ $\fn_jvn&space;=-3-5\sqrt{3}+5\sqrt{3}+25$ $\fn_jvn&space;=-3+25$ $\fn_jvn&space;=22$ ( মূলদ সংখ্যা )   উত্তরঃ $\fn_jvn&space;{\color{DarkGreen}&space;\left&space;({\color{DarkGreen}&space;-\sqrt{3}-5}&space;\right&space;)}$ [ যেহেতু $\fn_jvn&space;{\color{Blue}&space;\left&space;(&space;\sqrt{3}-5&space;\right&space;)}$  ও $\fn_jvn&space;{\color{Blue}&space;\left&space;(-\sqrt{3}-5&space;\right&space;)}$ এর যোগফল এবং গুনফল উভয়ই মূলদ সংখ্যা ]   (iii) দুটি দ্বিঘাত করণীর যোগফল ও গুনফল একটি মূলদ সংখ্যা হলে, করণীদ্বয় ________ করণী। উত্তরঃ অনুবন্ধী

 Q11. সংক্ষিপ্তধর্মী উত্তরপ্রশ্ন (S.A.) : (i)  $\small&space;{\color{Blue}&space;x=3+2\sqrt{2}}$  হলে,  $\small&space;{\color{Blue}&space;x+\frac{1}{x}}$  -এর মান লিখি। সমাধানঃ  $\fn_jvn&space;x=3+2\sqrt{2}$ $\fn_jvn&space;\therefore&space;\frac{1}{x}=\frac{1}{3+2\sqrt{2}}$ $\fn_jvn&space;=\frac{\left&space;(&space;(3-2\sqrt{2}&space;\right&space;)}{\left&space;(3+2\sqrt{2}&space;\right&space;)\left&space;(&space;(3-2\sqrt{2}&space;\right&space;)}$ [ হরের করণী নিরসন করে পাই ] $\fn_jvn&space;=\frac{\left&space;(&space;3-2\sqrt{2}&space;\right&space;)}{\left&space;[\left&space;(3&space;\right&space;)^{2}&space;\right&space;]-\left&space;[\left&space;(2\sqrt{2}&space;\right&space;)^{2}&space;\right&space;]&space;}$ $\fn_jvn&space;{\color{Blue}&space;\left&space;[&space;\because&space;\left&space;(&space;a+b&space;\right&space;)\left&space;(&space;a-b&space;\right&space;)=a^{2}-b^{2}&space;\right&space;]}$ $\fn_jvn&space;=\frac{\left&space;(&space;3-2\sqrt{2}&space;\right&space;)}{9-8}$ $\fn_jvn&space;=\frac{\left&space;(&space;3-2\sqrt{2}&space;\right&space;)}{1}$ $\fn_jvn&space;=\left&space;(&space;3-2\sqrt{2}&space;\right&space;)$ $\fn_jvn&space;\therefore&space;x+\frac{1}{x}$ $\fn_jvn&space;=3+2\sqrt{2}+3-2\sqrt{2}&space;{\color{Blue}&space;\left&space;[&space;\because&space;x=3+2\sqrt{2},\frac{1}{x}=3-2\sqrt{2}&space;\right&space;]}$ $\fn_jvn&space;=6$ উত্তরঃ নির্ণেয় $\fn_jvn&space;{\color{DarkGreen}&space;\left&space;(x+\frac{1}{x}&space;\right&space;)}$ এর মান $\fn_jvn&space;{\color{DarkGreen}&space;6}$ .

 Q11. সংক্ষিপ্তধর্মী উত্তরপ্রশ্ন (S.A.) : (ii)  $\small&space;{\color{Blue}&space;\left&space;(&space;\sqrt{15}+\sqrt{3}&space;\right&space;)}$  এবং  $\small&space;{\color{Blue}&space;\left&space;(&space;\sqrt{10}+\sqrt{8}&space;\right&space;)}$  এর মধ্যে কোনটি বড় লিখি। সমাধানঃ  $\fn_jvn&space;\left&space;(&space;\sqrt{15}+\sqrt{3}&space;\right&space;)^{2}$ [ বর্গ করে পাই ] $\fn_jvn&space;=15+2\sqrt{45}+3$ $\fn_jvn&space;=18+2\sqrt{3\times&space;3\times&space;5}$ $\fn_jvn&space;=18+6\sqrt{5}$ আবার,   $\fn_jvn&space;\left&space;(&space;\sqrt{10}+\sqrt{8}&space;\right&space;)^{2}$  [ বর্গ করে পাই ] $\fn_jvn&space;=10+2\sqrt{80}+8$ $\fn_jvn&space;=18+2\sqrt{5\times&space;4\times&space;4}$ $\fn_jvn&space;=18+8\sqrt{5}$ $\fn_jvn&space;\because&space;8\sqrt{5}>6\sqrt{5}$ $\fn_jvn&space;\therefore&space;\left&space;(&space;\sqrt{10}+\sqrt{8}&space;\right&space;)^{2}>\left(&space;\sqrt{15}+\sqrt{3}&space;\right)^{2}$ বা, $\fn_jvn&space;\left&space;(&space;\sqrt{10}+\sqrt{8}&space;\right&space;)>\left(&space;\sqrt{15}+\sqrt{3}&space;\right)$ উত্তরঃ নির্ণেয় $\fn_jvn&space;{\color{DarkGreen}&space;\left&space;(&space;\sqrt{15}+\sqrt{3}&space;\right&space;)}$  এবং  $\fn_jvn&space;{\color{DarkGreen}&space;\left&space;(&space;\sqrt{10}+\sqrt{8}&space;\right&space;)}$ এর মধ্যে $\fn_jvn&space;{\color{DarkGreen}&space;\left&space;(&space;\sqrt{10}+\sqrt{8}&space;\right&space;)}$  বড়ো।

 Q11. সংক্ষিপ্তধর্মী উত্তরপ্রশ্ন (S.A.) : (iii) দুটি মিশ্র দ্বিঘাত করণী লিখি যাদের গুণফল একটি মূলদ সংখ্যা। সমাধানঃ  যদি দুটি মিশ্র দ্বিঘাত করণী সদৃশ্য হয়, তবে তাদের গুনফল হবে। যেমন,  $\fn_jvn&space;2\sqrt{3}$ ও $\fn_jvn&space;5\sqrt{3}$ হলো দুটি মিশ্র সদৃশ্য দ্বিঘাত করণী যাদের গুনফল নিম্নরুপঃ  $\fn_jvn&space;2\sqrt{3}\times&space;5\sqrt{3}$ $\fn_jvn&space;=\left&space;(&space;2\times&space;5&space;\right&space;)\times&space;\left&space;(&space;\sqrt{3}\times&space;\sqrt{3}&space;\right&space;)$ $\fn_jvn&space;=10\times&space;\left&space;(&space;3&space;\right&space;)\;&space;\;&space;\;&space;{\color{Blue}&space;\left&space;[&space;\because&space;\sqrt{a}\times&space;\sqrt{a}=a&space;\right&space;]}$ $\fn_jvn&space;=&space;30$ … এটি হলো একটি মূলদ সংখ্যা।  উত্তরঃ নির্ণেয় মিশ্র করণী দুটি হলো $\fn_jvn&space;{\color{DarkGreen}&space;2\sqrt{3}}$ ও $\fn_jvn&space;{\color{DarkGreen}&space;5\sqrt{3}}$

 Q11. সংক্ষিপ্তধর্মী উত্তরপ্রশ্ন (S.A.) : (iv)  √72  থেকে কত বিয়োগ করলে √32  হবে তা লিখি। সমাধানঃ  ধরি,   $\fn_jvn&space;\sqrt{72}$ থেকে $\fn_jvn&space;x$ বিয়োগ করলে $\fn_jvn&space;\sqrt{32}$  হবে।  $\fn_jvn&space;\therefore&space;\sqrt{72}-x=\sqrt{32}$ বা, $\fn_jvn&space;6\sqrt{2}-x=4\sqrt{2}$ বা, $\fn_jvn&space;6\sqrt{2}-4\sqrt{2}=x$ $\fn_jvn&space;\therefore&space;x=2\sqrt{2}$ উত্তরঃ নির্ণেয় $\fn_jvn&space;{\color{DarkGreen}&space;\sqrt{72}}$ থেকে $\fn_jvn&space;{\color{DarkGreen}&space;2\sqrt{2}}$ বিয়োগ করলে $\fn_jvn&space;{\color{DarkGreen}&space;\sqrt{32}}$  হবে।

 Q11. সংক্ষিপ্তধর্মী উত্তরপ্রশ্ন (S.A.) : (v)  $\small&space;{\color{Blue}&space;\left[&space;\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}&space;\right]}$ – এর সরলতম মান লিখি। সমাধানঃ    $\fn_jvn&space;\left[&space;\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}&space;\right]$ $\fn_jvn&space;=\left[&space;\frac{\left&space;(\sqrt{2}-1&space;\right&space;)}{\left&space;(\sqrt{2}+1&space;\right&space;)\left&space;(\sqrt{2}-1&space;\right&space;)}+\frac{\left&space;(\sqrt{3}-\sqrt{2}&space;\right&space;)}{\left&space;(\sqrt{3}+\sqrt{2}&space;\right&space;)\left&space;(\sqrt{3}-\sqrt{2}&space;\right&space;)}+\frac{\left&space;(\sqrt{4}-\sqrt{3}&space;\right&space;)}{\left&space;(\sqrt{4}+\sqrt{3}&space;\right&space;)\left&space;(\sqrt{4}-\sqrt{3}&space;\right&space;)}&space;\right]$ [ হরের করণী নিরসন করে পাই ] $\fn_jvn&space;=\frac{\left&space;(&space;\sqrt{2}-1&space;\right&space;)}{\left&space;[\left&space;(\sqrt{2}&space;\right&space;)^{2}&space;\right&space;]-\left&space;[\left&space;(1&space;\right&space;)^{2}&space;\right&space;]&space;}+\frac{\left&space;(&space;\sqrt{3}-\sqrt{2}&space;\right&space;)}{\left&space;[\left&space;(\sqrt{3}&space;\right&space;)^{2}&space;\right&space;]-\left&space;[\left&space;(\sqrt{2}\right&space;)^{2}&space;\right&space;]&space;}+\frac{\left&space;(&space;\sqrt{4}-\sqrt{3}&space;\right&space;)}{\left&space;[\left&space;(\sqrt{4}&space;\right&space;)^{2}&space;\right&space;]-\left&space;[\left&space;(\sqrt{3}\right&space;)^{2}&space;\right&space;]&space;}$ $\fn_jvn&space;{\color{Blue}&space;\left&space;[&space;\because&space;\left&space;(&space;a+b&space;\right&space;)\left&space;(&space;a-b&space;\right&space;)=a^{2}-b^{2}&space;\right&space;]}$ $\fn_jvn&space;=\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}$ $\fn_jvn&space;=\frac{\sqrt{2}-1}{1}+\frac{\sqrt{3}-\sqrt{2}}{1}+\frac{\sqrt{4}-\sqrt{3}}{1}$ $\fn_jvn&space;=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}$ $\fn_jvn&space;=\sqrt{4}-1$ $\fn_jvn&space;=2-1$ $\fn_jvn&space;=1$ উত্তরঃ নির্ণেয়   $\fn_jvn&space;{\color{DarkGreen}&space;\left[&space;\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}&space;\right]}$ – এর সরলতম মান $\fn_jvn&space;{\color{DarkGreen}&space;1}$ .

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##### 6 thoughts on “Koshe dekhi 9.3 class 10”

Thank you for your valuable comment.

কি অসুবিধা Q8 এ?

2. অভিরূপ সরকার says:

খুব খুব ধন্যবাদ স্যার

#### Koshe dekhi 20.3 Class 8

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