Koshe dekhi 9.3 class 10

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

(i) $\fn_cm {\color{Blue} {{m}^{2}}+\frac{1}{{{m}^{2}}}}$

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

এদের সরলতম মান নির্ণয় করি।

Table of Contents

সমাধান (i) :

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

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

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

$\fn_cm =3-2$

$\fn_cm =1$

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

সমাধান (ii) :

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

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

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

$\fn_cm =3\sqrt{3}-3\sqrt{3}$

$\fn_cm =0$

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

Q1. (b) দেখাই যে, $\fn_cm \small {\color{Blue} \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=2\sqrt{15}}$

সমাধানঃ

বামপক্ষ $\fn_cm \small =\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$

$\fn_cm \small =\frac{\left ( \sqrt{5}+\sqrt{3} \right )^{2}-\left ( \sqrt{5}-\sqrt{3} \right )^{2}}{\left ( \sqrt{5}-\sqrt{3} \right )\left ( \sqrt{5}+\sqrt{3} \right )}$

$\small =\frac{\left [\left ( \sqrt{5} \right )^{2}+2\times \sqrt{5}\times \sqrt{3}+\left ( \sqrt{3} \right )^{2} \right ]-\left [\left ( \sqrt{5} \right )^{2}-2\times \sqrt{5}\times \sqrt{3}+\left ( \sqrt{3} \right )^{2} \right ]}{\left ( \sqrt{5} \right )^{2}-\left ( \sqrt{3} \right )^{2}}$ $\fn_cm {\color{Blue} \left [ \because \left ( a+b \right )\left ( a-b \right )=a^{2}-b^{2} \right ]}$

$\fn_cm =\frac{5+2\sqrt{15}+3-5+2\sqrt{15}-3}{5-3}$

$\fn_cm =\frac{4\sqrt{15}}{2}$

$\fn_cm =2\sqrt{15}=$ ডানপক্ষ ( প্রমাণিত )

Q2. সরল করি :

(a) $\fn_cm {\color{Blue} \frac{\sqrt{2}\left( 2+\sqrt{3} \right)}{\sqrt{3}\left( \sqrt{3}+1 \right)}-\frac{\sqrt{2}\left( 2-\sqrt{3} \right)}{\sqrt{3}\left( \sqrt{3}-1 \right)}}$

সমাধানঃ

$\fn_cm \frac{\sqrt{2}\left( 2+\sqrt{3} \right)}{\sqrt{3}\left( \sqrt{3}+1 \right)}-\frac{\sqrt{2}\left( 2-\sqrt{3} \right)}{\sqrt{3}\left( \sqrt{3}-1 \right)}$

$\fn_cm =\frac{\sqrt{2}}{\sqrt{3}}\left [\frac{\left( 2+\sqrt{3} \right)}{\left( \sqrt{3}+1 \right)}-\frac{\left( 2-\sqrt{3} \right)}{\left( \sqrt{3}-1 \right)} \right ]$

$\fn_cm =\frac{\sqrt{2}}{\sqrt{3}}\left [\frac{\left ( 2+\sqrt{3} \right )\left ( \sqrt{3}-1 \right )-\left ( 2-\sqrt{3} \right )\left ( \sqrt{3}+1 \right )}{\left ( \sqrt{3}+1 \right )\left ( \sqrt{3}-1 \right )} \right ]$

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

$\fn_cm =\frac{\sqrt{2}}{\sqrt{3}}\left [ \frac{2\sqrt{3}-2+3-\sqrt{3}-2\sqrt{3}-2+3+\sqrt{3}}{3-1} \right ]$

$\fn_phv =\frac{\sqrt{2}}{\sqrt{3}}\left [ \frac{2}{2} \right ]$

$\fn_cm =\frac{\sqrt{2}}{\sqrt{3}}$

$\fn_cm =\frac{\sqrt{2}\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}$ [ হরের করণী নিরসন করে পাই ]

$\fn_cm =\frac{\sqrt{6}}{3}$

উত্তরঃ নির্ণেয় সরলতম মান $\fn_cm {\color{DarkGreen} \frac{\sqrt{6}}{3}}$

Q2. সরল করি :

(b) $\fn_cm {\color{Blue} \frac{3\sqrt{7}}{\sqrt{5}+\sqrt{2}}-\frac{5\sqrt{5}}{\sqrt{2}+\sqrt{7}}+\frac{2\sqrt{2}}{\sqrt{7}+\sqrt{5}}}$

সমাধানঃ

$\fn_cm \frac{3\sqrt{7}}{\sqrt{5}+\sqrt{2}}-\frac{5\sqrt{5}}{\sqrt{2}+\sqrt{7}}+\frac{2\sqrt{2}}{\sqrt{7}+\sqrt{5}}$

$\fn_cm =\frac{3\sqrt{7}\left ( \sqrt{5}-\sqrt{2} \right )}{\left ( \sqrt{5}+\sqrt{2} \right )\left ( \sqrt{5}-\sqrt{2} \right )}-\frac{5\sqrt{5}\left ( \sqrt{2}-\sqrt{7} \right )}{\left ( \sqrt{2}+\sqrt{7} \right )\left ( \sqrt{2}-\sqrt{7} \right )}+\frac{2\sqrt{2}\left ( \sqrt{7}-\sqrt{5} \right )}{\left ( \sqrt{7}+\sqrt{5} \right )\left ( \sqrt{7}-\sqrt{5} \right )}$ [হরের করণী নিরসন করে পাই ]

$\fn_cm =\frac{3\sqrt{7}\left ( \sqrt{5}-\sqrt{2} \right )}{\left ( \sqrt{5} \right )^{2}-\left ( \sqrt{2} \right )^{2}}-\frac{5\sqrt{5}\left ( \sqrt{2}-\sqrt{7} \right )}{\left ( \sqrt{2} \right )^{2}-\left ( \sqrt{7} \right )^{2}}+\frac{2\sqrt{2}\left ( \sqrt{7}-\sqrt{5} \right )}{\left ( \sqrt{7} \right )^{2}-\left ( \sqrt{5} \right )^{2}}$

$=\frac{3\sqrt{7}\left ( \sqrt{5}-\sqrt{2} \right )}{5-2}-\frac{5\sqrt{5}\left ( \sqrt{2}-\sqrt{7} \right )}{2-7}+\frac{2\sqrt{2}\left ( \sqrt{7}-\sqrt{5} \right )}{7-5}$

$=\frac{3\sqrt{7}\left ( \sqrt{5}-\sqrt{2} \right )}{3}-\frac{5\sqrt{5}\left ( \sqrt{2}-\sqrt{7} \right )}{-5}+\frac{2\sqrt{2}\left ( \sqrt{7}-\sqrt{5} \right )}{2}$

$\fn_cm =\sqrt{7}\left ( \sqrt{5}-\sqrt{2} \right )+\sqrt{5}\left ( \sqrt{2}-\sqrt{7} \right )+\sqrt{2}\left ( \sqrt{7}-\sqrt{5} \right )$

$\fn_cm =\sqrt{35}-\sqrt{14}+\sqrt{10}-\sqrt{35}+\sqrt{14}-\sqrt{10}$

$\fn_cm =0$

উত্তরঃ নির্ণেয় সরলতম মান $\fn_jvn {\color{DarkGreen} 0}$

Q2. সরল করি :

(c) $\fn_cm {\color{Blue} \frac{4\sqrt{3}}{2-\sqrt{2}}-\frac{30}{4\sqrt{3}-\sqrt{18}}-\frac{\sqrt{18}}{3-\sqrt{12}}}$

সমাধানঃ

$\fn_cm \frac{4\sqrt{3}}{2-\sqrt{2}}-\frac{30}{4\sqrt{3}-\sqrt{18}}-\frac{\sqrt{18}}{3-\sqrt{12}}$

$\fn_cm =\frac{4\sqrt{3}\left ( 2+\sqrt{2} \right )}{\left ( 2-\sqrt{2} \right )\left ( 2+\sqrt{2} \right )}-\frac{30\left ( 4\sqrt{3}+\sqrt{18} \right )}{\left ( 4\sqrt{3}-\sqrt{18} \right )\left ( 4\sqrt{3}+\sqrt{18} \right )}-\frac{\sqrt{18}\left ( 3+\sqrt{12} \right )}{\left ( 3-\sqrt{12} \right )\left ( 3+\sqrt{12} \right )}$ [ হরের করণী নিরসন করে পাই ]

$\fn_cm =\frac{4\sqrt{3}\left ( 2+\sqrt{2} \right )}{\left ( 2 \right )^{2}-\left ( \sqrt{2} \right )^{2}}-\frac{30\left ( 4 \sqrt{3}+\sqrt{18} \right )}{\left ( 4\sqrt{3} \right )^{2}-\left ( \sqrt{18} \right )^{2}}-\frac{\sqrt{18}\left ( 3+\sqrt{12} \right )}{\left ( 3 \right )^{2}-\left ( \sqrt{12} \right )^{2}}\; {\color{Blue} \left [ \because \left ( a+b \right )\left ( a-b \right )=a^{2}-b^{2} \right ]}$

$\fn_cm =\frac{4\sqrt{3}\left ( 2+\sqrt{2} \right )}{4-2}-\frac{30\left ( 4 \sqrt{3}+\sqrt{18} \right )}{48-18}-\frac{\sqrt{18}\left ( 3+\sqrt{12} \right )}{9-12}$

$\fn_cm =\frac{4\sqrt{3}\left ( 2+\sqrt{2} \right )}{2}-\frac{30\left ( 4\sqrt{3}+3\sqrt{2} \right )}{30}-\frac{3\sqrt{2}\left ( 3+2\sqrt{3} \right )}{-3}\; {\color{Blue} \left [ \because \sqrt{18}=3\sqrt{2} ,\; \sqrt{12}=2\sqrt{3}\right ]}$

$\fn_cm =2\sqrt{3}\left ( 2+\sqrt{2} \right )-\left ( 4\sqrt{3}+3\sqrt{2} \right )+\sqrt{2}\left ( 3+2\sqrt{3} \right )$

$\fn_cm =4\sqrt{6}+2\sqrt{6}-4\sqrt{3}-3\sqrt{2}+3\sqrt{2}+2\sqrt{6}$

$\fn_cm =4\sqrt{6}$

উত্তরঃ নির্ণেয় সরলতম মান $\fn_cm {\color{DarkGreen} 4\sqrt{6}}$

Koshe dekhi 9.3 class 10

Q2. সরল করি :

(d) $\fn_cm {\color{Blue} \frac{3\sqrt{2}}{\sqrt{3}+\sqrt{6}}-\frac{4\sqrt{3}}{\sqrt{6}+\sqrt{2}}+\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}}$

সমাধানঃ

$\fn_cm \frac{3\sqrt{2}}{\sqrt{3}+\sqrt{6}}-\frac{4\sqrt{3}}{\sqrt{6}+\sqrt{2}}+\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}$

$\fn_cm =\frac{3\sqrt{2}\left (\sqrt{3}-\sqrt{6} \right )}{\left (\sqrt{3}+\sqrt{6} \right )\left (\sqrt{3}-\sqrt{6} \right )}-\frac{4\sqrt{3}\left (\sqrt{6}-\sqrt{2} \right )}{\left (\sqrt{6}+\sqrt{2} \right )\left (\sqrt{6}-\sqrt{2} \right )}+\frac{\sqrt{6}\left (\sqrt{2}-\sqrt{3} \right )}{\left (\sqrt{2}+\sqrt{3} \right )\left (\sqrt{2}-\sqrt{3} \right )}$ [ হরের করণী নিরসন করে পাই ]

$\fn_cm =\frac{3\sqrt{2}\left (\sqrt{3}-\sqrt{6} \right )}{\left ( \sqrt{3} \right )^{2}-\left ( \sqrt{6} \right )^{2}}-\frac{4\sqrt{3}\left ( \sqrt{6}-\sqrt{2} \right )}{\left ( \sqrt{6} \right )^{2}-\left ( \sqrt{2} \right )^{2}}+\frac{\sqrt{6}\left ( \sqrt{2}-\sqrt{3} \right )}{\left ( \sqrt{2} \right )^{2}-\left ( \sqrt{3} \right )^{2}}\; {\color{Blue} \left [ \because \left ( a+b \right )\left ( a-b \right )=a^{2}-b^{2} \right ]}$

$\fn_cm =\frac{3\sqrt{2}\left (\sqrt{3}-\sqrt{6} \right )}{3-6}-\frac{4\sqrt{3}\left ( \sqrt{6}-\sqrt{2} \right )}{6-2}+\frac{\sqrt{6}\left ( \sqrt{2}-\sqrt{3} \right )}{2-3}$

$\fn_cm =\frac{3\sqrt{2}\left (\sqrt{3}-\sqrt{6} \right )}{-3}-\frac{4\sqrt{3}\left ( \sqrt{6}-\sqrt{2} \right )}{4}+\frac{\sqrt{6}\left ( \sqrt{2}-\sqrt{3} \right )}{-1}$

$\fn_cm =-\sqrt{2}\left ( \sqrt{3}-\sqrt{6} \right )-\sqrt{3}\left ( \sqrt{6}-\sqrt{2} \right )-\sqrt{6}\left ( \sqrt{2}-\sqrt{3} \right )$

$\fn_cm =-\sqrt{6}+\sqrt{12}-\sqrt{18}+\sqrt{6}-\sqrt{12}+\sqrt{18}$

$\fn_cm =0$

উত্তরঃ নির্ণেয় সরলতম মান 0

Koshe dekhi 9.3 class 10

Q3. যদি $\fn_cm {\color{Blue} x=2,y=3}$ এবং $\fn_cm {\color{Blue} z=6}$ হয়, তবে, $\small {\color{Blue} \frac{3\sqrt{x}}{\sqrt{y}+\sqrt{z}}-\frac{4\sqrt{y}}{\sqrt{z}+\sqrt{x}}+\frac{\sqrt{z}}{\sqrt{x}+\sqrt{y}}}$ এর মান হিসাব করে লিখি।

সমাধানঃ

$\fn_cm \frac{3\sqrt{x}}{\sqrt{y}+\sqrt{z}}-\frac{4\sqrt{y}}{\sqrt{z}+\sqrt{x}}+\frac{\sqrt{z}}{\sqrt{x}+\sqrt{y}}$

$\fn_cm =\frac{3\sqrt{2}}{\sqrt{3}+\sqrt{6}}-\frac{4\sqrt{3}}{\sqrt{6}+\sqrt{2}}+\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}$ [ x , y ,z এর মান বসিয়ে পাই ]

$\fn_cm =\frac{3\sqrt{2}\left (\sqrt{3}-\sqrt{6} \right )}{3-6}-\frac{4\sqrt{3}\left ( \sqrt{6}-\sqrt{2} \right )}{6-2}+\frac{\sqrt{6}\left ( \sqrt{2}-\sqrt{3} \right )}{2-3}$

$\fn_cm =\frac{3\sqrt{2}\left (\sqrt{3}-\sqrt{6} \right )}{-3}-\frac{4\sqrt{3}\left ( \sqrt{6}-\sqrt{2} \right )}{4}+\frac{\sqrt{6}\left ( \sqrt{2}-\sqrt{3} \right )}{-1}$

$\fn_cm =-\sqrt{2}\left ( \sqrt{3}-\sqrt{6} \right )-\sqrt{3}\left ( \sqrt{6}-\sqrt{2} \right )-\sqrt{6}\left ( \sqrt{2}-\sqrt{3} \right )$

$\fn_cm =-\sqrt{6}+\sqrt{12}-\sqrt{18}+\sqrt{6}-\sqrt{12}+\sqrt{18}$

$\fn_cm =0$

উত্তরঃ নির্ণেয় $\fn_cm {\color{DarkGreen} \frac{3\sqrt{x}}{\sqrt{y}+\sqrt{z}}-\frac{4\sqrt{y}}{\sqrt{z}+\sqrt{x}}+\frac{\sqrt{z}}{\sqrt{x}+\sqrt{y}}}$ এর মান $\fn_jvn {\color{DarkGreen} 0}$ .

Koshe dekhi 9.3 class 10

Q4. $\fn_cm {\color{Blue} x=\sqrt{7}+\sqrt{6}}$ হলে,

(i) $\fn_cm {\color{Blue} x-\frac{1}{x}}$

(ii) $\fn_cm {\color{Blue} x+\frac{1}{x}}$

(iii) $\fn_cm {\color{Blue} {{x}^{2}}+\frac{1}{{{x}^{2}}}}$

(iv) $\fn_cm {\color{Blue} {{x}^{3}}+\frac{1}{{{x}^{3}}}}$

এদের সরলতম মান নির্ণয় করি।

সমাধানঃ

$\fn_cm \because x=\left ( \sqrt{7} +\sqrt{6}\right )$

$\fn_cm \therefore \frac{1}{x}=\frac{1}{\left ( \sqrt{7}+\sqrt{6} \right )}$

$\fn_cm =\frac{1}{\left ( \sqrt{7}+\sqrt{6} \right )}\times \frac{\left ( \sqrt{7}-\sqrt{6} \right )}{\left ( \sqrt{7}-\sqrt{6} \right )}$ [ হরের করণী নিরসন করে পাই ]

$\fn_cm =\frac{\left ( \sqrt{7}-\sqrt{6} \right )}{\left ( \sqrt{7} \right )^{2}-\left ( \sqrt{6} \right )^{2}}$

$\fn_cm =\frac{\left ( \sqrt{7}-\sqrt{6} \right )}{7-6}$

$\fn_cm =\left ( \sqrt{7}-\sqrt{6} \right )$

সমাধান (i) :

এখন, (i) $\fn_cm x-\frac{1}{x}$

$\fn_jvn =\left ( \sqrt{7}+\sqrt{6} \right )-\left ( \sqrt{7}-\sqrt{6} \right )$ [ x ও $\fn_cm {\color{Blue} \frac{1}{x}}$ এর মান বসিয়ে পাই ]

$\fn_cm =\sqrt{7}+\sqrt{6}-\sqrt{7}+\sqrt{6}$

$\fn_cm =2\sqrt{6}$

উত্তরঃ নির্ণেয় $\fn_cm {\color{DarkGreen} x-\frac{1}{x}}$ এর সরলতম মান $\fn_cm {\color{DarkGreen} 2\sqrt{6}}$

সমাধান (ii) :

(ii) $\fn_cm x+\frac{1}{x}$

$\fn_cm =\left ( \sqrt{7}+\sqrt{6} \right )+\left ( \sqrt{7}-\sqrt{6} \right )$ [ x ও $\fn_cm {\color{Blue} \frac{1}{x}}$ এর মান বসিয়ে পাই ]

$\fn_cm =\sqrt{7}+\sqrt{6}+\sqrt{7}-\sqrt{6}$

$\fn_cm =2\sqrt{7}$

উত্তরঃ নির্ণেয় $\fn_cm {\color{DarkGreen} x+\frac{1}{x}}$ এর সরলতম মান $\fn_cm {\color{DarkGreen} 2\sqrt{7}}$

সমাধান (iii) :

(iii) $\fn_cm {{x}^{2}}+\frac{1}{{{x}^{2}}}$

$\fn_cm =\left ( x+\frac{1}{x} \right )^{2}-2\cdot x\cdot \frac{1}{x}$

$\fn_cm =\left ( 2\sqrt{7} \right )^{2}-2$ [ $\fn_cm {\color{Blue} x+\frac{1}{x}}$ এর মান বসিয়ে পাই ]

$\fn_cm =28-2$

$\fn_cm =26$

উত্তরঃ নির্ণেয় $\fn_cm {\color{DarkGreen} {{x}^{2}}+\frac{1}{{{x}^{2}}}}$ এর সরলতম মান $\fn_cm {\color{DarkGreen} 26}$

সমাধান (iv) :

(iv) $\fn_cm {{x}^{3}}+\frac{1}{{{x}^{3}}}$

$\fn_cm =\left ( x+\frac{1}{x} \right )^{3}-3\cdot x\cdot \frac{1}{x}\cdot \left ( x+\frac{1}{x} \right )$

$\fn_cm =\left ( 2\sqrt{7} \right )^{3}-3\times 2\sqrt{7}$ [ $\fn_cm {\color{Blue} x+\frac{1}{x}}$ এর মান বসিয়ে পাই ]

$\fn_cm =56\sqrt{7}-6\sqrt{7}$

$\fn_cm =50\sqrt{7}$

উত্তরঃ নির্ণেয় $\fn_cm {\color{DarkGreen} {{x}^{3}}+\frac{1}{{{x}^{3}}}}$ এর সরলতম মান $\fn_cm {\color{DarkGreen} 50\sqrt{7}}$ .

Koshe dekhi 9.3 class 10

Q5. সরল করি : $\small {\color{Blue} \frac{x+\sqrt{{{x}^{2}}-1}}{x-\sqrt{{{x}^{2}}-1}}+\frac{x-\sqrt{{{x}^{2}}-1}}{x+\sqrt{{{x}^{2}}-1}}}$ সরলফল $\fn_cm {\color{Blue} 14}$ হলে, $\fn_cm {\color{Blue} x}$ -এর মান কী কী হবে হিসাব করে লিখি।

সমাধানঃ

$\fn_cm \frac{x+\sqrt{{{x}^{2}}-1}}{x-\sqrt{{{x}^{2}}-1}}+\frac{x-\sqrt{{{x}^{2}}-1}}{x+\sqrt{{{x}^{2}}-1}}=14$

বা, $\fn_cm \frac{\left ( x+\sqrt{x^{2}-1} \right )^{2}+\left ( x-\sqrt{x^{2}-1} \right )^{2}}{\left ( x-\sqrt{x^{2}-1} \right )\left ( x+\sqrt{x^{2}-1} \right )}=14$

বা, $\fn_cm \frac{2\left [\left ( x\right )^{2} +\left ( \sqrt{x^{2}-1} \right ) ^{2}\right ]}{\left ( x\right )^{2} -\left ( \sqrt{x^{2}-1} \right ) ^{2}}=14$ $\fn_cm {\color{Blue}\left [\because \left ( i \right ) \left ( a+b \right )^{2}+\left ( a-b \right )^{2}=2\left (a^{2}+b ^{2} \right ), \left ( ii \right ) \left ( a-b \right )\left ( a+b \right )=a^{2}-b^{2}\right ]}$

বা, $\fn_cm \frac{2\left ( x^{2} + x^{2}-1\right )}{ x^{2}- x^{2}+1}=14$

বা, $\fn_cm \frac{2\left ( 2x^{2}-1 \right )}{1}=14$

বা, $\fn_cm 2x^{2}-1=\frac{14}{2}$

বা, $\fn_cm 2x^{2}=7+1$

বা, $\fn_cm 2x^{2}=8$

বা, $\fn_cm x^{2}=\frac{8}{2}=4$

বা, $\fn_cm x=\pm \sqrt{4}$

$\fn_cm \therefore x=\pm 2$

উত্তরঃ নির্ণেয় $\fn_cm {\color{DarkGreen} x}$ -এর মান $\fn_cm {\color{DarkGreen} \pm 2}$ হবে।

Koshe dekhi 9.3 class 10

Q6. যদি $\fn_cm {\color{Blue} a=\frac{\sqrt{5}+1}{\sqrt{5}-1}}$ ও $\fn_cm {\color{Blue} b=\frac{\sqrt{5}-1}{\sqrt{5}+1}}$ হয়, তবে নীচের মানগুলি নির্ণয় করি।

(i) $\fn_cm {\color{Blue} \frac{{{a}^{2}}+ab+{{b}^{2}}}{{{a}^{2}}-ab+{{b}^{2}}}}$

(ii) $\fn_cm {\color{Blue} \frac{{{\left( a-b \right)}^{3}}}{{{\left( a+b \right)}^{3}}}}$

(iii) $\fn_cm {\color{Blue} \frac{3{{a}^{2}}+5ab+3{{b}^{2}}}{3{{a}^{2}}-5ab+3{{b}^{2}}}}$

(iv) $\fn_cm {\color{Blue} \frac{{{a}^{3}}+{{b}^{3}}}{{{a}^{3}}-{{b}^{3}}}}$

সমাধানঃ

$\fn_cm a=\frac{\sqrt{5}+1}{\sqrt{5}-1}$ , $\fn_cm b=\frac{\sqrt{5}-1}{\sqrt{5}+1}$

∴ a + b

$\fn_cm =\frac{\sqrt{5}+1}{\sqrt{5}-1}+\frac{\sqrt{5}-1}{\sqrt{5}+1}$

$\fn_cm =\frac{\left ( \sqrt{5}+1 \right )^{2}+\left ( \sqrt{5}-1 \right )^{2}}{\left ( \sqrt{5}-1 \right )\left ( \sqrt{5}+1 \right )}$

$\fn_cm =\frac{2\left [ \left ( \sqrt{5} \right )^{2}+\left ( 1 \right )^{2} \right ]}{\left ( \sqrt{5} \right )^{2}-\left ( 1 \right )^{2}}$ $\fn_cm {\color{Blue}\left [\because \left ( i \right ) \left ( a+b \right )^{2}+\left ( a-b \right )^{2}=2\left (a^{2}+b ^{2} \right ), \left ( ii \right ) \left ( a-b \right )\left ( a+b \right )=a^{2}-b^{2}\right ]}$

$\fn_cm =\frac{2\left ( 5+1 \right )}{\left ( 5-1 \right )}$

$\fn_cm =\frac{2\times 6}{4}$

$\fn_cm =3$

এবং,

$\fn_cm \left ( a-b \right )$ $\fn_cm =\frac{\sqrt{5}+1}{\sqrt{5}-1}-\frac{\sqrt{5}-1}{\sqrt{5}+1}$

$\fn_cm =\frac{\left ( \sqrt{5}+1 \right )^{2}-\left ( \sqrt{5}-1 \right )^{2}}{\left ( \sqrt{5}-1 \right )\left ( \sqrt{5}+1 \right )}$

$\fn_cm =\frac{4\times \sqrt{5}\times 1}{\left ( \sqrt{5} \right )^{2}-\left ( 1 \right )^{2}}$ $\fn_cm {\color{Blue}\left [\because \left ( i \right ) \left ( a+b \right )^{2}-\left ( a-b \right )^{2}=4ab, \left ( ii \right ) \left ( a-b \right )\left ( a+b \right )=a^{2}-b^{2}\right ]}$

$\fn_cm =\frac{4\sqrt{5}}{\left ( 5-1 \right )}$

$\fn_cm =\frac{4\sqrt{5}}{4}$

$\fn_cm =\sqrt{5}$

আবার,

$\fn_cm ab=\frac{\left (\sqrt{5}+1 \right )}{\left (\sqrt{5}-1 \right )}\times \frac{\left (\sqrt{5}-1 \right )}{\left (\sqrt{5}+1 \right )}$

$\fn_cm \therefore ab=1$

সমাধান (i) :

এখন, (i) $\fn_cm \frac{{{a}^{2}}+ab+{{b}^{2}}}{{{a}^{2}}-ab+{{b}^{2}}}$

$\fn_cm =\frac{\left \{ \left ( a+b \right )^{2}-2ab \right \}+ab}{\left \{ \left ( a+b \right )^{2}-2ab \right \}-ab} {\color{Blue} \left [ \because a^{2}+b^{2}=\left ( a+b \right )^{2}-2ab \right ]}$

$\fn_cm =\frac{\left ( 3 \right )^{2}-2\times 1+1}{\left ( 3 \right )^{2}-2\times 1-1}$ [ $\fn_cm {\color{Blue} a+b=3}$ এবং $\fn_cm {\color{Blue} ab=1}$ বসিয়ে পাই ]

$\fn_cm =\frac{9-2+1}{9-2-1}$

$\fn_cm =\frac{8}{6}$

$\fn_cm =\frac{4}{3}$

$\fn_cm =1\frac{1}{3}$

উত্তরঃ নির্ণেয় $\fn_cm {\color{DarkGreen} \frac{{{a}^{2}}+ab+{{b}^{2}}}{{{a}^{2}}-ab+{{b}^{2}}}}$ এর মান $\fn_cm {\color{DarkGreen} 1\frac{1}{3}}$

সমাধান (ii) :

(ii) $\fn_cm \frac{{{\left( a-b \right)}^{3}}}{{{\left( a+b \right)}^{3}}}$

$\fn_cm =\frac{\left ( \sqrt{5} \right )^{3}}{\left ( 3 \right )^{3}}$ $\fn_cm {\color{Blue}[\because a-b=\sqrt{5}}$ এবং $\fn_cm {\color{Blue} a+b=3]}$

$\fn_cm =\frac{5\sqrt{5}}{27}$

উত্তরঃ নির্ণেয় $\fn_cm {\color{DarkGreen} \frac{{{\left( a-b \right)}^{3}}}{{{\left( a+b \right)}^{3}}}}$ এর মান $\fn_cm {\color{DarkGreen} \frac{5\sqrt{5}}{27}}$

সমাধান (iii) :

(iii) $\fn_cm \frac{3{{a}^{2}}+5ab+3{{b}^{2}}}{3{{a}^{2}}-5ab+3{{b}^{2}}}$

$\fn_cm =\frac{3\left ( a^{2} +b^{2} \right )+5ab}{3\left ( a^{2} +b^{2} \right )-5ab}$

$\fn_cm =\frac{3\left [\left ( a+b \right )^{2}-2ab \right ]+5ab}{3\left [\left ( a+b \right )^{2}-2ab \right ]-5ab} {\color{Blue} \left [ \because a^{2}+b^{2}=\left ( a+b \right )^{2}-2ab \right ]}$

$\fn_cm =\frac{3\left [\left ( 3 \right )^{2}-2\times 1 \right ]+1}{3\left [\left ( 3 \right )^{2}-2\times 1 \right ]-1}$ [ $\fn_cm {\color{Blue} a+b=3}$ এবং $\fn_cm {\color{Blue} ab=1}$ বসিয়ে পাই ]

$\fn_cm =\frac{3\left [ 9-2 \right ]+5\times 1}{3\left [ 9-2 \right ]-5\times 1}$

$\fn_cm =\frac{3\times 7+5}{3\times 7-5}$

$\fn_cm =\frac{26}{16}$

$\fn_cm =\frac{13}{8}$

$\fn_cm =1\frac{5}{8}$

উত্তরঃ নির্ণেয় $\fn_cm {\color{DarkGreen} \frac{3{{a}^{2}}+5ab+3{{b}^{2}}}{3{{a}^{2}}-5ab+3{{b}^{2}}}}$ এর মান $\fn_cm {\color{DarkGreen} 1\frac{5}{8}}$

সমাধান (iv) :

(iv) $\fn_cm \frac{{{a}^{3}}+{{b}^{3}}}{{{a}^{3}}-{{b}^{3}}}$

$\fn_cm =\frac{\left ( a+b \right )^{3}-3ab\left ( a+b \right )}{\left ( a-b \right )^{3}+3ab\left ( a-b \right )}$

$\fn_cm =\frac{\left ( 3 \right )^{3}-3\times 1\times 3}{\left ( \sqrt{5} \right )^{3}+3\times 1\times \sqrt{5}}$

$\fn_cm =\frac{27-9}{5\sqrt{5}+3\sqrt{5}}$

$\fn_cm =\frac{18}{8\sqrt{5}}$

$\fn_cm =\frac{9}{4\sqrt{5}}$

$\fn_cm =\frac{9\times \sqrt{5}}{4\sqrt{5}\times \sqrt{5}}$

$\fn_cm =\frac{9\sqrt{5}}{20}$

উত্তরঃ নির্ণেয় $\fn_cm {\color{DarkGreen} \frac{{{a}^{3}}+{{b}^{3}}}{{{a}^{3}}-{{b}^{3}}}}$ এর মান $\fn_cm {\color{DarkGreen} \frac{9\sqrt{5}}{20}}$

Koshe dekhi 9.3 class 10

Q7. যদি $\fn_cm {\color{Blue} x=2+\sqrt{3}}$ এবং $\fn_cm {\color{Blue} y=2-\sqrt{3}}$ হয়, তবে নিম্নলিখিতগুলির সরলতম মান নির্ণয় করি :

(a)

(i) $\fn_cm {\color{Blue} x-\frac{1}{x}}$

(ii) $\fn_cm {\color{Blue} {{y}^{2}}+\frac{1}{{{y}^{2}}}}$

(iii) $\fn_cm {\color{Blue} {{x}^{3}}-\frac{1}{{{x}^{3}}}}$

(iv) $\fn_cm {\color{Blue} xy+\frac{1}{xy}}$

(b) $\fn_cm {\color{Blue} 3{{x}^{2}}-5xy+3{{y}^{2}}}$

সমাধানঃ

$\fn_cm \because x=2+\sqrt{3}$

$\fn_cm \therefore \frac{1}{x}=\frac{1}{2+\sqrt{3}}$

বা, $\fn_cm \frac{1}{x}=\frac{\left (2-\sqrt{3} \right )}{\left (2+\sqrt{3} \right )\left (2-\sqrt{3} \right )}$ [ হরের করণী নিরসন করে পাই ]

বা, $\fn_cm \frac{1}{x}=\frac{2-\sqrt{3}}{\left ( 2 \right ) ^{2}-\left ( \sqrt{3} \right )^{2}}$ $\fn_cm {\color{Blue} \left [ \because \left ( a+b \right )\left ( a-b \right )=a^{2}-b^{2} \right ]}$

বা, $\fn_cm \frac{1}{x}=\frac{2-\sqrt{3}}{4-3}\Rightarrow 2-\sqrt{3}$

$\fn_cm \therefore \frac{1}{x}=2-\sqrt{3}$

একইভাবে,

$\fn_cm \because y=2-\sqrt{3}$

$\fn_cm \therefore \frac{1}{y}=\frac{1}{2-\sqrt{3}}$

বা, $\fn_cm \frac{1}{y}=\frac{\left (2+\sqrt{3} \right )}{\left (2+\sqrt{3} \right )\left (2-\sqrt{3} \right )}$ [ হরের করণী নিরসন করে পাই ]

বা, $\fn_cm \frac{1}{y}=\frac{2+\sqrt{3}}{\left ( 2 \right ) ^{2}-\left ( \sqrt{3} \right )^{2}}$ $\fn_cm {\color{Blue} \left [ \because \left ( a+b \right )\left ( a-b \right )=a^{2}-b^{2} \right ]}$

বা, $\fn_cm \frac{1}{y}=\frac{2+\sqrt{3}}{4-3}\Rightarrow 2+\sqrt{3}$

$\fn_cm \therefore \frac{1}{y}=2+\sqrt{3}$

সমাধান (i) :

এখন, (i) $\fn_cm {\color{Blue} x-\frac{1}{x}}$

$\fn_cm =\left ( 2+\sqrt{3} \right )-\left ( 2-\sqrt{3} \right )$ [ $\fn_cm {\color{Blue} x}$ ও $\fn_cm {\color{Blue} \frac{1}{x}}$ -এর মান বসিয়ে পাই ]

$\fn_cm =2+\sqrt{3}-2+\sqrt{3}$

$\fn_cm =2\sqrt{3}$

উত্তরঃ নির্ণেয় $\fn_cm {\color{DarkGreen} x-\frac{1}{x}}$ এর সরলতম মান $\fn_cm {\color{DarkGreen} 2\sqrt{3}}$ .

সমাধান (ii) :

(ii) $\fn_cm {\color{Blue} {{y}^{2}}+\frac{1}{{{y}^{2}}}}$

$\fn_cm \because y=2-\sqrt{3}$ এবং $\fn_cm \frac{1}{y}=2+\sqrt{3}$

$\fn_cm \therefore y+\frac{1}{y}$

$\fn_cm =\left ( 2-\sqrt{3} \right )+\left ( 2+\sqrt{3} \right )$ [ $\fn_cm {\color{Blue} y}$ ও $\fn_cm {\color{Blue} \frac{1}{y}}$ -এর মান বসিয়ে পাই ]

$\fn_cm =2-\sqrt{3}+2+\sqrt{3}$

$\fn_cm =4$

সুতরাং, $\fn_cm y+\frac{1}{y}=4$

এখন, $\fn_cm {{y}^{2}}+\frac{1}{{{y}^{2}}}$

$\fn_cm =\left ( y+\frac{1}{y} \right )^{2}-2\cdot y\cdot \frac{1}{y}$

$\fn_cm =4^{2}-2\; {\color{Blue} \left [\because y+\frac{1}{y}=4 \right ]}$

$\fn_cm =16-2\Rightarrow 14$

উত্তরঃ নির্ণেয় $\fn_cm {\color{DarkGreen} {{y}^{2}}+\frac{1}{{{y}^{2}}}}$ এর সরলতম মান $\fn_cm {\color{DarkGreen} 14}$ .

সমাধান (iii) :

(iii) $\fn_cm {\color{Blue} {{x}^{3}}-\frac{1}{{{x}^{3}}}}$

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

$\fn_cm =\left ( 2\sqrt{3} \right )^{3}+3\times 2\sqrt{3}$

$\fn_cm =24\sqrt{3}+6\sqrt{3}$

$\fn_cm =30\sqrt{3}$

উত্তরঃ নির্ণেয় $\fn_cm {\color{DarkGreen} {{x}^{3}}-\frac{1}{{{x}^{3}}}}$ এর সরলতম মান $\fn_cm {\color{DarkGreen} 30\sqrt{3}}$ .

সমাধান (iv) :

(iv) $\fn_cm {\color{Blue} xy+\frac{1}{xy}}$

$\fn_cm \because x=2+\sqrt{3}$ এবং $\fn_cm y=2-\sqrt{3}$

$\fn_cm \therefore xy=\left ( 2+\sqrt{3} \right )\left ( 2-\sqrt{3} \right )$

$\fn_cm =\left ( 2 \right )^{2}-\left ( 2\sqrt{3} \right )^{2}$

$\fn_cm =4-3$

$\fn_cm =1$

$\fn_cm \therefore xy+\frac{1}{xy}$

$\fn_cm =1+\frac{1}{1}$ [ $\fn_cm {\color{Blue} xy=1}$ বসিয়ে পাই ]

$\fn_cm =1+1$

$\fn_cm =2$

উত্তরঃ নির্ণেয় $\fn_cm {\color{DarkGreen} xy+\frac{1}{xy}}$ এর সরলতম মান 2.

সমাধান (b) :

(b) $\fn_cm {\color{Blue} 3{{x}^{2}}-5xy+3{{y}^{2}}}$

$\fn_cm \because x=2+\sqrt{3}$ এবং $\fn_cm y=2-\sqrt{3}$

$\fn_cm \therefore x+y=\left ( 2+\sqrt{3} \right )+\left ( 2-\sqrt{3} \right )$

$\fn_cm =2+\sqrt{3}+2-\sqrt{3}$

$\fn_cm =4$

এখন, $\fn_cm 3x^{2}-5xy+3y^{2}$

$\fn_cm =3\left ( x^{2}+y^{2} \right )-5xy$

$\fn_cm =3\left \{ \left ( x+y \right )^{2}-2xy \right \}-5xy$

$\fn_cm =3\left \{ \left ( 4 \right )^{2}-2\times 1 \right \}-5\times 1$ $\fn_cm {\color{Blue} \left [ \because x+y=4,\; xy=1 \right ]}$

$\fn_cm =3\left ( 16-2 \right )-5$

$\fn_cm =3\times 14-5\Rightarrow 42-5$

$\fn_cm =37$

উত্তরঃ নির্ণেয় $\fn_cm {\color{DarkGreen} 3{{x}^{2}}-5xy+3{{y}^{2}}}$ এর সরলতম মান $\fn_cm {\color{DarkGreen}37}.$

Koshe dekhi 9.3 class 10

Q8. $\small {\color{Blue} x=\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}}}$ এবং $\small {\color{Blue} xy=1}$ হলে, দেখাই যে, $\small {\color{Blue} \frac{{{x}^{2}}+xy+{{y}^{2}}}{{{x}^{2}}-xy+{{y}^{2}}}=\frac{12}{11}}$

সমাধানঃ

প্রদত্ত,

$\fn_cm \because x=\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}}$ এবং $\fn_cm xy=1$

$\fn_cm \therefore y=\frac{1}{x}$

$\fn_cm =\frac{1}{\left (\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}} \right )}$

$\fn_cm =\frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}}$

এবং, $\fn_cm x+y$

$\fn_cm =\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}}+\frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}}$ [ x ও y এর মান বসিয়ে পাই ]

$\fn_cm =\frac{\left ( \sqrt{7}+\sqrt{3} \right )^{2}+\left ( \sqrt{7}-\sqrt{3} \right )^{2}}{\left ( \sqrt{7}+\sqrt{3} \right )\left ( \sqrt{7}-\sqrt{3} \right )}$

$\fn_cm =\frac{2\left [ \left ( \sqrt{7} \right )^{2}+\left ( \sqrt{3} \right )^{2} \right ]}{\left ( \sqrt{7} \right )^{2}-\left ( \sqrt{3} \right )^{2}}$ $\fn_cm {\color{Blue}\left [\because \left ( i \right ) \left ( a+b \right )^{2}+\left ( a-b \right )^{2}=2\left (a^{2}+b ^{2} \right ), \left ( ii \right ) \left ( a-b \right )\left ( a+b \right )=a^{2}-b^{2}\right ]}$

$\fn_cm =\frac{2\left ( 7+3 \right )}{\left ( 7-3 \right )}$

$\fn_cm =\frac{20}{4}=5$

$\fn_cm \therefore x+y=5$

বামপক্ষ:

$\fn_cm \frac{{{x}^{2}}+xy+{{y}^{2}}}{{{x}^{2}}-xy+{{y}^{2}}}$

$\fn_cm =\frac{\left ( x^{2}+y^{2} \right )+xy}{\left ( x^{2}-y^{2} \right )-xy}$

$\fn_cm =\frac{\left [ \left ( x+y \right )^{2}-2xy \right ]+xy}{\left [ \left ( x+y \right )^{2}-2xy \right ]-xy}$

$\fn_cm =\frac{\left ( 5 \right )^{2}-2\times 1+1}{\left ( 5 \right )^{2}-2\times 1-1}$ $\fn_cm {\color{Blue} \left [ \because x+y=5,xy=1 \right ]}$

$\fn_cm =\frac{25-2+1}{25-2-1}$

$\fn_cm =\frac{24}{22}$

$\fn_cm =\frac{12}{11}=$ ডানপক্ষ ( প্রমাণিত )

Koshe dekhi 9.3 class 10

Q9. $\fn_cm {\color{Blue} \left( \sqrt{7}+1 \right)}$ এবং $\fn_cm {\color{Blue} \left( \sqrt{5}+\sqrt{3} \right)}$ এর মধ্যে কোনটি বড়ো লিখি।

সমাধানঃ

$\fn_jvn \left( \sqrt{7}+1 \right)^{2}$ [ বর্গ করে পাই ]

$\fn_jvn =7+2\sqrt{7}+1$

$\fn_jvn =8+2\sqrt{7}$

আবার,

$\fn_jvn \left( \sqrt{5}+\sqrt{3} \right)^{2}$ [ বর্গ করে পাই ]

$\fn_jvn =5+2\sqrt{15}+3$

$\fn_jvn =8+2\sqrt{15}$

$\fn_jvn \because \sqrt{15}>\sqrt{7}$

$\fn_jvn \therefore \left ( \sqrt{5}+\sqrt{3} \right )^{2}>\left( \sqrt{7}+1 \right)^{2}$

বা, $\fn_jvn \left ( \sqrt{5}+\sqrt{3} \right )>\left( \sqrt{7}+1 \right)$

উত্তরঃ নির্ণেয় $\fn_jvn {\color{DarkGreen} \left( \sqrt{7}+1 \right)}$ এবং $\fn_jvn {\color{DarkGreen} \left( \sqrt{5}+\sqrt{3} \right)}$ এর মধ্যে $\fn_jvn {\color{DarkGreen} \left( \sqrt{5}+\sqrt{3} \right)}$ বড়ো।

Koshe dekhi 9.3 class 10

Q10. অতিরিক্ত সংক্ষিপ্ত উত্তরধর্মী প্রশ্ন (V.S.A.)

(A) বহুবিকল্পীয় প্রশ্ন (M.C.Q.)

(i) $\small {\color{Blue} x=2+\sqrt{3}}$ হলে, $\small {\color{Blue} x+\frac{1}{x}}$ এর মান –

(a) 2

(b) 2√3

(d) 2 − √3

সমাধানঃ

$\fn_jvn x=2+\sqrt{3}$ ,

$\fn_jvn \frac{1}{x}=\frac{1}{2+\sqrt{3}}$

বা, $\fn_jvn \frac{1}{x}=\frac{\left ( (2-\sqrt{3} \right )}{\left (2+\sqrt{3} \right )\left (2-\sqrt{3} \right )}$ [ হরের করণী নিরসন করে পাই ]

বা, $\fn_jvn \frac{1}{x}=\frac{2-\sqrt{3}}{\left [ \left ( 2 \right ) ^{2}-\left ( \sqrt{3} \right )^{2}\right ]}$ $\fn_jvn {\color{Blue} \left [ \because \left ( a+b \right )\left ( a-b \right )=a^{2}-b^{2} \right ]}$

$\fn_jvn \therefore \frac{1}{x}=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}$

$\fn_jvn \therefore x+\frac{1}{x}$

$\fn_jvn =2+\sqrt{3}+2-\sqrt{3}$

$\fn_jvn =4$

উত্তরঃ $\fn_jvn {\color{DarkGreen} \left ( c \right )4}$

Koshe dekhi 9.3 class 10

Q10. অতিরিক্ত সংক্ষিপ্ত উত্তরধর্মী প্রশ্ন (V.S.A.)

(A) বহুবিকল্পীয় প্রশ্ন (M.C.Q.)

(ii) যদি $\small {\color{Blue} p+q=\sqrt{13}}$ এবং $\small {\color{Blue} p-q=\sqrt{5}}$ হয়, তাহলে $\small {\color{Blue} pq}$ এর মান –

(a) 2

(b) 18

(d) 8

সমাধানঃ

$\fn_jvn p+q=\sqrt{13}$

বা, $\fn_jvn \left ( p+q \right )^{2}=\left ( \sqrt{13} \right )^{2}$ [ উভয়পক্ষে বর্গ করে পাই ]

বা, $\fn_jvn \left ( p+q \right )^{2}=13......\left ( i \right )$

আবার,

$\fn_jvn p-q=\sqrt{5}$

বা, $\fn_jvn \left ( p-q \right )^{2}=\left ( \sqrt{5} \right )^{2}$ [ উভয়পক্ষে বর্গ করে পাই ]

বা, $\fn_jvn \left ( p-q \right )^{2}=5......\left ( ii \right )$

$\fn_jvn \left ( i \right )$ নং সমীকরণ থেকে $\fn_jvn \left ( ii \right )$ নং সমীকরণ বিয়োগ করে পাই

$\fn_jvn \left ( p+q \right )^{2}-\left ( p-q \right )^{2}=13-5$

বা, $\fn_jvn 4pq=8 {\color{Blue} \left [ \because \left ( a+b \right )^{2}-\left ( a-b \right )^{2}=4ab \right ]}$

$\fn_jvn \therefore pq=\frac{8}{4}=2$

উত্তরঃ $\fn_jvn {\color{DarkGreen} \left ( a \right )2}$

Koshe dekhi 9.3 class 10

Q10. অতিরিক্ত সংক্ষিপ্ত উত্তরধর্মী প্রশ্ন (V.S.A.)

(A) বহুবিকল্পীয় প্রশ্ন (M.C.Q.)

(iii) যদি $\small {\color{Blue} a+b=\sqrt{5}}$ এবং $\small {\color{Blue} a-b=\sqrt{3}}$ তাহলে $\small {\color{Blue} \left( {{a}^{2}}+{{b}^{2}} \right)}$ এর মান –

(a) 8

(b) 4

(d) 1

সমাধানঃ

$\fn_jvn a+b=\sqrt{5}$

বা, $\fn_jvn \left ( a+b \right )^{2}=\left ( \sqrt{5} \right )^{2}$ [ উভয়পক্ষে বর্গ করে পাই ]

বা, $\fn_jvn \left ( a+b \right )^{2}=5......\left ( i \right )$

আবার,

$\fn_jvn a-b=\sqrt{3}$

বা, $\fn_jvn \left ( a-b \right )^{2}=\left ( \sqrt{3} \right )^{2}$ [ উভয়পক্ষে বর্গ করে পাই ]

বা, $\fn_jvn \left ( a-b \right )^{2}=3......\left ( ii \right )$

$\fn_jvn \left ( i \right )$ নং সমীকরণ ও $\fn_jvn \left ( ii \right )$ নং সমীকরণ যোগ করে পাই

$\fn_jvn \left ( a+b \right )^{2}+\left ( a-b \right )^{2}=5+3$

বা, $\fn_jvn a^{2}+2ab+b^{2}+a^{2}-2ab+b^{2}=8$

বা, $\fn_jvn 2\left( {{a}^{2}}+{{b}^{2}} \right)=8$

$\fn_jvn \therefore \left( {{a}^{2}}+{{b}^{2}} \right)=\frac{8}{2}=4$

উত্তরঃ $\fn_jvn {\color{DarkGreen} \left ( b \right )4}$

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

(d) কোনটিই নয়

সমাধানঃ

$\fn_jvn \sqrt{125}$ থেকে $\fn_jvn \sqrt{5}$ বিয়োগ করলে বিয়োগফল হবে-

$\fn_jvn \sqrt{125}-\sqrt{5}$

$\fn_jvn =\sqrt{5\times 5\times 5}-\sqrt{5}$

$\fn_jvn =5\sqrt{5}-\sqrt{5}$

$\fn_jvn =4\sqrt{5}$

$\fn_jvn =\sqrt{4\times 4\times 5}$

$\fn_jvn =\sqrt{80}$

উত্তরঃ $\fn_jvn {\color{DarkGreen} \left ( a \right )\sqrt{80}}$

Koshe dekhi 9.3 class 10

Q10. অতিরিক্ত সংক্ষিপ্ত উত্তরধর্মী প্রশ্ন (V.S.A.)

(A) বহুবিকল্পীয় প্রশ্ন (M.C.Q.)

(v) $\small {\color{Blue} \left( 5-\sqrt{3} \right)\left( \sqrt{3}-1 \right)\left( 5+\sqrt{3} \right)\left( \sqrt{3}+1 \right)}$ এর গুণফল –

(a) 22

(b) 44

(d) 11

সমাধানঃ

$\fn_jvn \left( 5-\sqrt{3} \right)\left( \sqrt{3}-1 \right)\left( 5+\sqrt{3} \right)\left( \sqrt{3}+1 \right)$ এর গুণফল –

$\fn_jvn \left( 5-\sqrt{3} \right)\times \left( \sqrt{3}-1 \right)\times \left( 5+\sqrt{3} \right)\times \left( \sqrt{3}+1 \right)$

$\fn_jvn =\left [\left( 5-\sqrt{3} \right)\times \left( 5+\sqrt{3} \right) \right ]\times\left [ \left( \sqrt{3}-1 \right)\times \left( \sqrt{3}+1 \right) \right ]$

$\fn_jvn =\left [\left (5 \right )^{2}-\left (\sqrt{3} \right )^{2} \right ]\times\left [ \left( \sqrt{3}\right)^{2}-\left ( 1 \right )^{2} \right ] {\color{Blue} \left [ \because \left ( a-b \right )\left ( a+b \right )=a^{2}-b^{2} \right ]}$

$\fn_jvn =\left [ 25-3 \right ]\times \left [ 3-1 \right ]$

$\fn_jvn =22\times 2$

$\fn_jvn =44$

উত্তরঃ $\fn_jvn {\color{DarkGreen} \left ( b \right )44}$

Koshe dekhi 9.3 class 10

(B) নীচের বিবৃতিগুলি সত্য না মিথ্যা লিখি :

(i) $\fn_jvn \sqrt{75}$ এবং $\fn_jvn \sqrt{147}$ সদৃশ করণী।

সমাধানঃ

$\fn_jvn \sqrt{75}$

$\fn_jvn =\sqrt{5\times 5\times 3}$

$\fn_jvn =5\sqrt{3}$

এবং

$\fn_jvn \sqrt{147}$

$\fn_jvn =\sqrt{7\times 7\times 3}$

$\fn_jvn =7\sqrt{3}$

∴ $\fn_jvn \sqrt{75}$ এবং $\fn_jvn \sqrt{147}$ সদৃশ করণী।

উত্তরঃ বিবৃতিটি সত্য।

(ii) $\fn_jvn \sqrt{\pi}$ একটি দ্বিঘাত করণী।

সমাধানঃ

বিবৃতিটি মিথ্যা।

কারণ : যে সমস্ত সংখ্যাকে $\fn_jvn \pm \sqrt{a}$ আকারে লেখা যায়, যেখানে $\fn_jvn a$ হলো ধনাত্মক পূর্ণসংখ্যা।

যেহেতু, $\fn_jvn \pi$ পূর্ণসংখ্যা নয় ( $\fn_jvn \pi$ হলো অমূলদ সংখ্যা ), তাই $\fn_jvn \sqrt{\pi}$ দ্বিঘাত করোনি নয়।

উত্তরঃ বিবৃতিটি মিথ্যা।

(C) শূন্যস্থান পূরণ করি :

(i) $\fn_jvn 5\sqrt{11}$ একটি _______ সংখ্যা। (মূলদ / অমূলদ)

উত্তরঃ অমূলদ [ যেহেতু $\fn_jvn {\color{Blue} 5\sqrt{11}}$ সংখ্যাটির সুনির্দিষ্ট দশমিক মান সম্পূর্ণভাবে নির্ণয় করা যায় না, তাই এটি একটি অমূলদ সংখ্যা ]

(ii) $\fn_jvn \left ( \sqrt{3}-5 \right )$ এর অনুবন্ধী করণী _______ ।

সমাধানঃ

$\fn_jvn \left ( \sqrt{3}-5 \right )$ ও $\fn_jvn \left (-\sqrt{3}-5 \right )$ এর যোগফল

$\fn_jvn \left ( \sqrt{3}-5 \right )+\left (- \sqrt{3}-5 \right )$

$\fn_jvn =-10$ ( মূলদ সংখ্যা )

$\fn_jvn \left ( \sqrt{3}-5 \right )$ ও $\fn_jvn \left (-\sqrt{3}-5 \right )$ এর গুনফল

$\fn_jvn \left ( \sqrt{3}-5 \right )\times \left (- \sqrt{3}-5 \right )$

$\fn_jvn =-3-5\sqrt{3}+5\sqrt{3}+25$

$\fn_jvn =-3+25$

$\fn_jvn =22$ ( মূলদ সংখ্যা )

উত্তরঃ $\fn_jvn {\color{DarkGreen} \left ({\color{DarkGreen} -\sqrt{3}-5} \right )}$ [ যেহেতু $\fn_jvn {\color{Blue} \left ( \sqrt{3}-5 \right )}$ ও $\fn_jvn {\color{Blue} \left (-\sqrt{3}-5 \right )}$ 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) $\small {\color{Blue} x=3+2\sqrt{2}}$ হলে, $\small {\color{Blue} x+\frac{1}{x}}$ -এর মান লিখি।

সমাধানঃ

$\fn_jvn x=3+2\sqrt{2}$

$\fn_jvn \therefore \frac{1}{x}=\frac{1}{3+2\sqrt{2}}$

$\fn_jvn =\frac{\left ( (3-2\sqrt{2} \right )}{\left (3+2\sqrt{2} \right )\left ( (3-2\sqrt{2} \right )}$ [ হরের করণী নিরসন করে পাই ]

$\fn_jvn =\frac{\left ( 3-2\sqrt{2} \right )}{\left [\left (3 \right )^{2} \right ]-\left [\left (2\sqrt{2} \right )^{2} \right ] }$ $\fn_jvn {\color{Blue} \left [ \because \left ( a+b \right )\left ( a-b \right )=a^{2}-b^{2} \right ]}$

$\fn_jvn =\frac{\left ( 3-2\sqrt{2} \right )}{9-8}$

$\fn_jvn =\frac{\left ( 3-2\sqrt{2} \right )}{1}$

$\fn_jvn =\left ( 3-2\sqrt{2} \right )$

$\fn_jvn \therefore x+\frac{1}{x}$

$\fn_jvn =3+2\sqrt{2}+3-2\sqrt{2} {\color{Blue} \left [ \because x=3+2\sqrt{2},\frac{1}{x}=3-2\sqrt{2} \right ]}$

$\fn_jvn =6$

উত্তরঃ নির্ণেয় $\fn_jvn {\color{DarkGreen} \left (x+\frac{1}{x} \right )}$ এর মান $\fn_jvn {\color{DarkGreen} 6}$ .

Q11. সংক্ষিপ্তধর্মী উত্তরপ্রশ্ন (S.A.) :

(ii) $\small {\color{Blue} \left ( \sqrt{15}+\sqrt{3} \right )}$ এবং $\small {\color{Blue} \left ( \sqrt{10}+\sqrt{8} \right )}$ এর মধ্যে কোনটি বড় লিখি।

সমাধানঃ

$\fn_jvn \left ( \sqrt{15}+\sqrt{3} \right )^{2}$ [ বর্গ করে পাই ]

$\fn_jvn =15+2\sqrt{45}+3$

$\fn_jvn =18+2\sqrt{3\times 3\times 5}$

$\fn_jvn =18+6\sqrt{5}$

আবার,

$\fn_jvn \left ( \sqrt{10}+\sqrt{8} \right )^{2}$ [ বর্গ করে পাই ]

$\fn_jvn =10+2\sqrt{80}+8$

$\fn_jvn =18+2\sqrt{5\times 4\times 4}$

$\fn_jvn =18+8\sqrt{5}$

$\fn_jvn \because 8\sqrt{5}>6\sqrt{5}$

$\fn_jvn \therefore \left ( \sqrt{10}+\sqrt{8} \right )^{2}>\left( \sqrt{15}+\sqrt{3} \right)^{2}$

বা, $\fn_jvn \left ( \sqrt{10}+\sqrt{8} \right )>\left( \sqrt{15}+\sqrt{3} \right)$

উত্তরঃ নির্ণেয় $\fn_jvn {\color{DarkGreen} \left ( \sqrt{15}+\sqrt{3} \right )}$ এবং $\fn_jvn {\color{DarkGreen} \left ( \sqrt{10}+\sqrt{8} \right )}$ এর মধ্যে $\fn_jvn {\color{DarkGreen} \left ( \sqrt{10}+\sqrt{8} \right )}$ 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, $\fn_jvn 2\sqrt{3}$ ও $\fn_jvn 5\sqrt{3}$ হলো দুটি মিশ্র সদৃশ্য দ্বিঘাত করণী যাদের গুনফল নিম্নরুপঃ

$\fn_jvn 2\sqrt{3}\times 5\sqrt{3}$

$\fn_jvn =\left ( 2\times 5 \right )\times \left ( \sqrt{3}\times \sqrt{3} \right )$

$\fn_jvn =10\times \left ( 3 \right )\; \; \; {\color{Blue} \left [ \because \sqrt{a}\times \sqrt{a}=a \right ]}$

$\fn_jvn = 30$ … এটি হলো একটি মূলদ সংখ্যা।

উত্তরঃ নির্ণেয় মিশ্র করণী দুটি হলো $\fn_jvn {\color{DarkGreen} 2\sqrt{3}}$ ও $\fn_jvn {\color{DarkGreen} 5\sqrt{3}}$

Q11. সংক্ষিপ্তধর্মী উত্তরপ্রশ্ন (S.A.) :

(iv) √72 থেকে কত বিয়োগ করলে √32 হবে তা লিখি।

সমাধানঃ

ধরি, $\fn_jvn \sqrt{72}$ থেকে $\fn_jvn x$ বিয়োগ করলে $\fn_jvn \sqrt{32}$ হবে।

$\fn_jvn \therefore \sqrt{72}-x=\sqrt{32}$

বা, $\fn_jvn 6\sqrt{2}-x=4\sqrt{2}$

বা, $\fn_jvn 6\sqrt{2}-4\sqrt{2}=x$

$\fn_jvn \therefore x=2\sqrt{2}$

উত্তরঃ নির্ণেয় $\fn_jvn {\color{DarkGreen} \sqrt{72}}$ থেকে $\fn_jvn {\color{DarkGreen} 2\sqrt{2}}$ বিয়োগ করলে $\fn_jvn {\color{DarkGreen} \sqrt{32}}$ হবে।

Q11. সংক্ষিপ্তধর্মী উত্তরপ্রশ্ন (S.A.) :

(v) $\small {\color{Blue} \left[ \frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}} \right]}$ – এর সরলতম মান লিখি।

সমাধানঃ

$\fn_jvn \left[ \frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}} \right]$

$\fn_jvn =\left[ \frac{\left (\sqrt{2}-1 \right )}{\left (\sqrt{2}+1 \right )\left (\sqrt{2}-1 \right )}+\frac{\left (\sqrt{3}-\sqrt{2} \right )}{\left (\sqrt{3}+\sqrt{2} \right )\left (\sqrt{3}-\sqrt{2} \right )}+\frac{\left (\sqrt{4}-\sqrt{3} \right )}{\left (\sqrt{4}+\sqrt{3} \right )\left (\sqrt{4}-\sqrt{3} \right )} \right]$ [ হরের করণী নিরসন করে পাই ]

$\fn_jvn =\frac{\left ( \sqrt{2}-1 \right )}{\left [\left (\sqrt{2} \right )^{2} \right ]-\left [\left (1 \right )^{2} \right ] }+\frac{\left ( \sqrt{3}-\sqrt{2} \right )}{\left [\left (\sqrt{3} \right )^{2} \right ]-\left [\left (\sqrt{2}\right )^{2} \right ] }+\frac{\left ( \sqrt{4}-\sqrt{3} \right )}{\left [\left (\sqrt{4} \right )^{2} \right ]-\left [\left (\sqrt{3}\right )^{2} \right ] }$ $\fn_jvn {\color{Blue} \left [ \because \left ( a+b \right )\left ( a-b \right )=a^{2}-b^{2} \right ]}$

$\fn_jvn =\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}$

$\fn_jvn =\frac{\sqrt{2}-1}{1}+\frac{\sqrt{3}-\sqrt{2}}{1}+\frac{\sqrt{4}-\sqrt{3}}{1}$

$\fn_jvn =\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}$

$\fn_jvn =\sqrt{4}-1$

$\fn_jvn =2-1$

$\fn_jvn =1$

উত্তরঃ নির্ণেয় $\fn_jvn {\color{DarkGreen} \left[ \frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}} \right]}$ – এর সরলতম মান $\fn_jvn {\color{DarkGreen} 1}$ .