Wed. Aug 7th, 2024

Feb 1, 2020

# Koshe Dekhi 5.3 class 10

## Koshe Dekhi 5.3 class 10

 Q1.  a : b = c : d  হলে, দেখাই যে, (i) ${\color{Blue}&space;\left(&space;{{a}^{2}}+{{b}^{2}}&space;\right):\left(&space;{{a}^{2}}-{{b}^{2}}&space;\right)=\left(&space;ac+bd&space;\right):\left(&space;ac-bd&space;\right)}$ সমাধানঃ প্রদত্ত,  a : b = c : d এখন,  $\frac{a}{b}=\frac{c}{d}=k$ (ধরি) $\therefore&space;a=bk,c=dk$ বামপক্ষ : $\left(&space;{{a}^{2}}+{{b}^{2}}&space;\right):\left(&space;{{a}^{2}}-{{b}^{2}}&space;\right)$ $=\frac{\left(&space;{{a}^{2}}+{{b}^{2}}&space;\right)}{\left(&space;{{a}^{2}}-{{b}^{2}}&space;\right)}$ a  এর মান বসিয়ে পাই, $=\frac{\left&space;(&space;bk&space;\right&space;)^{2}+b^{2}}{\left&space;(&space;bk&space;\right&space;)^{2}-b^{2}}$ $=\frac{b^{2}k^{2}+b^{2}}{b^{2}k^{2}-b^{2}}$ $=\frac{b^{2}\left&space;(k^{2}+1&space;\right&space;)}{b^{2}\left&space;(k^{2}-1&space;\right&space;)}$ $=\frac{\left&space;(k^{2}+1&space;\right&space;)}{\left&space;(k^{2}-1&space;\right&space;)}$ ডানপক্ষ : $\left&space;(&space;ac+bd&space;\right&space;):\left&space;(&space;ac-bd&space;\right&space;)$ $=\frac{ac+bd}{ac-bd}$ a  ও   c  এর মান বসিয়ে পাই, $=\frac{bk\times&space;dk+bd}{bk\times&space;dk-bd}$ $=\frac{bdk^{2}+bd}{bdk^{2}-bd}$ $=\frac{bd\left&space;(k^{2}+1&space;\right&space;)}{bd\left&space;(k^{2}-1&space;\right&space;)}$ $=\frac{\left&space;(k^{2}+1&space;\right&space;)}{\left&space;(k^{2}-1&space;\right&space;)}$ ∴ বামপক্ষ = ডানপক্ষ (প্রমাণিত)

 Q1 (ii) ${\color{Blue}&space;\left(&space;{{a}^{2}}+ab+{{b}^{2}}&space;\right):\left(&space;{{a}^{2}}-ab+{{b}^{2}}&space;\right)=\left(&space;{{c}^{2}}+cd+{{d}^{2}}&space;\right):\left(&space;{{c}^{2}}-cd+{{d}^{2}}&space;\right)}$ সমাধানঃ প্রদত্ত,  a : b = c : d এখন,  $\frac{a}{b}=\frac{c}{d}=k$ (ধরি) $\therefore&space;a=bk,c=dk$ বামপক্ষ : $\left(&space;{{a}^{2}}+ab+{{b}^{2}}&space;\right):\left(&space;{{a}^{2}}-ab+{{b}^{2}}&space;\right)$ $=\frac{\left(&space;{{a}^{2}}+ab+{{b}^{2}}&space;\right)}{\left(&space;{{a}^{2}}-ab+{{b}^{2}}&space;\right)}$ a এর মান বসিয়ে পাই, $=\frac{\left&space;(&space;bk&space;\right&space;)^{2}+bk\times&space;b+b^{2}}{\left&space;(&space;bk&space;\right&space;)^{2}-bk\times&space;b+b^{2}}$ $=\frac{b^{2}k^{2}+b^{2}k+b^{2}}{b^{2}k^{2}-b^{2}k+b^{2}}$ $=\frac{b^{2}\left&space;(k^{2}+k+1&space;\right&space;)}{b^{2}\left&space;(k^{2}-k+1&space;\right&space;)}$ $=\frac{\left&space;(k^{2}+k+1&space;\right&space;)}{\left&space;(k^{2}-k+1&space;\right&space;)}$ ডানপক্ষ : $\left(&space;{{c}^{2}}+cd+{{d}^{2}}&space;\right):\left(&space;{{c}^{2}}-cd+{{d}^{2}}&space;\right)$ $=\frac{\left(&space;{{c}^{2}}+cd+{{d}^{2}}&space;\right)}{\left(&space;{{c}^{2}}-cd+{{d}^{2}}&space;\right)}$ c  এর মান বসিয়ে পাই, $=\frac{\left&space;(&space;dk&space;\right&space;)^{2}+dk\times&space;d+d^{2}}{\left&space;(&space;dk&space;\right&space;)^{2}-dk\times&space;d+d^{2}}$ $=\frac{d^{2}k^{2}+d^{2}k+d^{2}}{d^{2}k^{2}-d^{2}k+d^{2}}$ $=\frac{d^{2}\left&space;(k^{2}+k+1&space;\right&space;)}{d^{2}\left&space;(k^{2}-k+1&space;\right&space;)}$ $=\frac{\left&space;(k^{2}+k+1&space;\right&space;)}{\left&space;(k^{2}-k+1&space;\right&space;)}$ ∴ বামপক্ষ = ডানপক্ষ (প্রমাণিত)

 Q1 (iii) ${\color{Blue}&space;\sqrt{{{a}^{2}}+{{c}^{2}}}:\sqrt{{{b}^{2}}+{{d}^{2}}}=\left(&space;pa+qc&space;\right):\left(&space;pb+qd&space;\right)}$ সমাধানঃ প্রদত্ত,  a : b = c : d এখন,  $\frac{a}{b}=\frac{c}{d}=k$ (ধরি) $\therefore&space;a=bk,c=dk$ বামপক্ষ : $\sqrt{{{a}^{2}}+{{c}^{2}}}:\sqrt{{{b}^{2}}+{{d}^{2}}}$ $=\frac{\sqrt{{{a}^{2}}+{{c}^{2}}}}{\sqrt{{{b}^{2}}+{{d}^{2}}}}$ a ও c এর মান বসিয়ে পাই, $=\frac{\sqrt{\left&space;(&space;bk&space;\right&space;)^{2}+\left&space;(&space;dk&space;\right&space;)^{2}}}{\sqrt{b^{2}+d^{2}}}$ $=\frac{\sqrt{b^{2}k^{2}+d^{2}k^{2}}}{\sqrt{b^{2}+d^{2}}}$ $=\frac{\sqrt{k^{2}\left&space;(b^{2}+d^{2}&space;\right&space;)}}{\sqrt{\left&space;(b^{2}+d^{2}&space;\right&space;)}}$ $=\sqrt{k^{2}}$ = k ডানপক্ষ : $\left(&space;pa+qc&space;\right):\left(&space;pb+qd&space;\right)$ $=\frac{\left(&space;pa+qc&space;\right)}{\left(&space;pb+qd&space;\right)}$ a ও c এর মান বসিয়ে পাই, $=\frac{\left(&space;p\times&space;bk+q\times&space;dk&space;\right)}{\left(&space;pb+qd&space;\right)}$ $=\frac{\left(&space;pbk+qdk&space;\right)}{\left(&space;pb+qd&space;\right)}$ $=\frac{k\left(&space;pb+qd&space;\right)}{\left(&space;pb+qd&space;\right)}$ = k ∴ বামপক্ষ = ডানপক্ষ (প্রমাণিত)

 Q2.  x : a = y : b = z : c  হলে, প্রমান করি যে, (i) ${\color{Blue}&space;\frac{{{x}^{3}}}{{{a}^{2}}}+\frac{{{y}^{3}}}{{{b}^{2}}}+\frac{{{z}^{3}}}{{{c}^{2}}}=\frac{{{\left(&space;x+y+z&space;\right)}^{3}}}{{{\left(&space;a+b+c&space;\right)}^{2}}}}$ সমাধানঃ প্রদত্ত, x : a = y : b = z : c এখন, $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k$ (ধরি) $\therefore&space;x=ak,y=bk,z=ck$ বামপক্ষ : $\frac{{{x}^{3}}}{{{a}^{2}}}+\frac{{{y}^{3}}}{{{b}^{2}}}+\frac{{{z}^{3}}}{{{c}^{2}}}$ x , y ও z এর মান বসিয়ে পাই, $=\frac{{{\left&space;(ak&space;\right&space;)}^{3}}}{{{a}^{2}}}+\frac{{{\left&space;(&space;bk&space;\right&space;)}^{3}}}{{{b}^{2}}}+\frac{{{\left&space;(&space;ck&space;\right&space;)}^{3}}}{{{c}^{2}}}$ $=\frac{{a^{3}k^{3}}}{{{a}^{2}}}+\frac{{b^{3}k^{3}}}{{{b}^{2}}}+\frac{{c^{3}k^{3}}}{{{c}^{2}}}$ $=ak^{3}+bk^{3}+ck^{3}$ $=k^{3}\left&space;(&space;a+b+c&space;\right&space;)$ ডানপক্ষ : $\frac{{{\left(&space;x+y+z&space;\right)}^{3}}}{{{\left(&space;a+b+c&space;\right)}^{2}}}$ x , y ও z এর মান বসিয়ে পাই, $=\frac{{{\left(&space;ak+bk+ck&space;\right)}^{3}}}{{{\left(&space;a+b+c&space;\right)}^{2}}}$ $=\frac{{{k^{3}\left(&space;a+b+c&space;\right)}^{3}}}{{{\left(&space;a+b+c&space;\right)}^{2}}}$ $=k^{3}\left&space;(&space;a+b+c&space;\right&space;)$ ∴ বামপক্ষ = ডানপক্ষ (প্রমাণিত)

 Q2.  x : a = y : b = z : c  হলে, প্রমান করি যে, (ii) ${\color{Blue}&space;\frac{{{x}^{3}}+{{y}^{3}}+{{z}^{3}}}{{{a}^{3}}+{{b}^{3}}+{{c}^{3}}}=\frac{xyz}{abc}}$ সমাধানঃ $\frac{{{x}^{3}}+{{y}^{3}}+{{z}^{3}}}{{{a}^{3}}+{{b}^{3}}+{{c}^{3}}}=\frac{xyz}{abc}$ প্রদত্ত, x : a = y : b = z : c এখন, $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k$ (ধরি) $\therefore&space;x=ak,y=bk,z=ck$ বামপক্ষ : $\frac{{{x}^{3}}+{{y}^{3}}+{{z}^{3}}}{{{a}^{3}}+{{b}^{3}}+{{c}^{3}}}$ x , y ও z এর মান বসিয়ে পাই, $=\frac{{{\left&space;(ak&space;\right&space;)}^{3}}+{{\left&space;(&space;bk&space;\right&space;)}^{3}}+{\left&space;(&space;ck&space;\right&space;)}^{3}}{a^{3}+b^{3}+c^{3}}$ $=\frac{{a^{3}k^{3}}+{b^{3}k^{3}}+{c^{3}k^{3}}}{a^{3}+b^{3}+c^{3}}$ $=\frac{k^{3}\left&space;({a^{3}}+{b^{3}}+{c^{3}}&space;\right&space;)}{a^{3}+b^{3}+c^{3}}$ $=k^{3}$ ডানপক্ষ :   $\frac{xyz}{abc}$ x , y ও z এর মান বসিয়ে পাই, $=\frac{ak\times&space;bk\times&space;ck}{a\times&space;b\times&space;c}$ $=\frac{abck^{3}}{abc}$ $=k^{3}$ ∴ বামপক্ষ = ডানপক্ষ (প্রমাণিত)

 Q2.  x : a = y : b = z : c  হলে, প্রমান করি যে, (iii) ${\color{Blue}&space;\left(&space;{{a}^{2}}+{{b}^{2}}+{{c}^{2}}&space;\right)\left(&space;{{x}^{2}}+{{y}^{2}}+{{z}^{2}}&space;\right)={{\left(&space;ax+by+cz&space;\right)}^{2}}}$ সমাধানঃ $\left(&space;{{a}^{2}}+{{b}^{2}}+{{c}^{2}}&space;\right)\left(&space;{{x}^{2}}+{{y}^{2}}+{{z}^{2}}&space;\right)={{\left(&space;ax+by+cz&space;\right)}^{2}}$ প্রদত্ত, x : a = y : b = z : c এখন, $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k$ (ধরি) $\therefore&space;x=ak,y=bk,z=ck$ বামপক্ষ : $\left(&space;{{a}^{2}}+{{b}^{2}}+{{c}^{2}}&space;\right)\left(&space;{{x}^{2}}+{{y}^{2}}+{{z}^{2}}&space;\right)$ x , y ও z এর মান বসিয়ে পাই, $=\left(&space;{{a}^{2}}+{{b}^{2}}+{{c}^{2}}&space;\right)&space;\left&space;[{{\left&space;(ak&space;\right&space;)}^{2}}+{{\left&space;(&space;bk&space;\right&space;)}^{2}}+{{\left&space;(&space;ck&space;\right&space;)}^{2}}&space;\right&space;]$ $=\left(&space;{{a}^{2}}+{{b}^{2}}+{{c}^{2}}&space;\right)&space;\left&space;[{a^{2}k^{2}}+{b^{2}k&space;^{2}}+{c^{2}k^{2}}&space;\right&space;]$ $=\left(&space;{{a}^{2}}+{{b}^{2}}+{{c}^{2}}&space;\right)&space;\left&space;[k^{2}\left&space;({a^{2}}+{b^{2}}+{c^{2}}&space;\right&space;)&space;\right&space;]$ $=k^{2}\left(&space;{{a}^{2}}+{{b}^{2}}+{{c}^{2}}&space;\right)^{2}$ ডানপক্ষ : ${{\left(&space;ax+by+cz&space;\right)}^{2}}$ x , y ও z এর মান বসিয়ে পাই, $={{\left(&space;a\times&space;ak+b\times&space;bk+c\times&space;ck&space;\right)}^{2}}$ $=\left(&space;{{a}^{2}k}+{{b}^{2}k}+{{c}^{2}k}&space;\right)^{2}$ $=k^{2}\left(&space;{{a}^{2}}+{{b}^{2}}+{{c}^{2}}&space;\right)^{2}$ ∴ বামপক্ষ = ডানপক্ষ (প্রমাণিত)

 Q3.  a : b = c : d = e : f  হলে, প্রমান করি যে, (i) প্রত্যেকটি অনুপাত ${\color{Blue}&space;=\frac{5a-7c-13e}{5b-7d-13f}}$ সমাধানঃ প্রদত্ত, a : b = c : d = e : f  বা, $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=k$  (ধরি) $\therefore&space;a=bk,c=dk,e=fk$   $\frac{5a-7c-13e}{5b-7d-13f}$ a , b ও c এর মান বসিয়ে পাই, $=\frac{5\times&space;bk-7\times&space;dk-13\times&space;fk}{5b-7d-13f}$ $=\frac{5bk-7dk-13fk}{5b-7d-13f}$ $=\frac{k\left&space;(5b-7d-13f&space;\right&space;)}{5b-7d-13f}$ $=k$ $=\frac{a}{b}=\frac{c}{d}=\frac{e}{f}$ = প্রত্যেকটি অনুপাত (প্রমাণিত)

 Q3.  a : b = c : d = e : f  হলে, প্রমান করি যে, (ii) ${\color{Blue}&space;\left(&space;{{a}^{2}}+{{c}^{2}}+{{e}^{2}}&space;\right)\left(&space;{{b}^{2}}+{{d}^{2}}+{{f}^{2}}&space;\right)={{\left(&space;ab+cd+ef&space;\right)}^{2}}}$ সমাধানঃ প্রদত্ত, a : b = c : d = e : f  বা, $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=k$  (ধরি) $\therefore&space;a=bk,c=dk,e=fk$ বামপক্ষ : $\left(&space;{{a}^{2}}+{{c}^{2}}+{{e}^{2}}&space;\right)\left(&space;{{b}^{2}}+{{d}^{2}}+{{f}^{2}}&space;\right)$ a , b ও c এর মান বসিয়ে পাই, $=\left[&space;{{\left&space;(bk&space;\right&space;)}^{2}}+{{\left&space;(dk&space;\right&space;)}^{2}}+{{\left&space;(fk&space;\right&space;)}^{2}}&space;\right]\left(&space;{{b}^{2}}+{{d}^{2}}+{{f}^{2}}&space;\right)$ $=\left[&space;{b^{2}k^{2}}+{d^{2}k^{2}}+{f^{2}k^{2}}&space;\right]\left(&space;{{b}^{2}}+{{d}^{2}}+{{f}^{2}}&space;\right)$ $=\left[&space;k^{2}\left(&space;{{b}^{2}}+{{d}^{2}}+{{f}^{2}}&space;\right)&space;\right]\left(&space;{{b}^{2}}+{{d}^{2}}+{{f}^{2}}&space;\right)$ $=k^{2}\left(&space;{{b}^{2}}+{{d}^{2}}+{{f}^{2}}&space;\right)^{2}$ ডানপক্ষ : ${{\left(&space;ab+cd+ef&space;\right)}^{2}}$ a , b ও c এর মান বসিয়ে পাই, $={{\left(&space;bk\times&space;b+dk\times&space;d+fk\times&space;f&space;\right)}^{2}}$ $={{\left(&space;b^{2}k+d^{2}k+f^{2}k&space;\right)}^{2}}$ $=k^{2}\left(&space;{{b}^{2}}+{{d}^{2}}+{{f}^{2}}&space;\right)^{2}$ ∴ বামপক্ষ = ডানপক্ষ (প্রমাণিত)

 Q4. যদি  a : b = b : c  হয়, তবে প্রমান করি যে, (i) ${\color{Blue}&space;{{\left(&space;\frac{a+b}{b+c}&space;\right)}^{2}}=\frac{{{a}^{2}}+{{b}^{2}}}{{{b}^{2}}+{{c}^{2}}}}$ সমাধানঃ a : b = b : c বা, $\frac{a}{b}=\frac{b}{c}=k$ (ধরি) $\therefore&space;b=ck,a=bk=ck\times&space;k=ck^{2}$ বামপক্ষ : ${{\left(&space;\frac{a+b}{b+c}&space;\right)}^{2}}$ a ও b এর মান বসিয়ে পাই, $={{\left(&space;\frac{ck^{2}+ck}{ck+c}&space;\right)}^{2}}$ $={c^{2}{\left(&space;\frac{k^{2}+k}{k+1}&space;\right)}^{2}}$ ডানপক্ষ : $\frac{{{a}^{2}}+{{b}^{2}}}{{{b}^{2}}+{{c}^{2}}}$ a ও b এর মান বসিয়ে পাই, $=\frac{{{\left&space;(ck^{2}&space;\right&space;)}^{2}}+{{\left&space;(&space;ck&space;\right&space;)}^{2}}}{{{\left&space;(&space;ck&space;\right&space;)}^{2}}+{{c}^{2}}}$ $=\frac{{{\left&space;(c^{2}k^{4}&space;\right&space;)}}+{{\left&space;(&space;c^{2}k^{2}&space;\right&space;)}}}{{{\left&space;(&space;c^{2}k^{2}&space;\right&space;)}}+{{c}^{2}}}$ $={c^{2}{\left(&space;\frac{k^{4}+k^{2}}{k^{2}+1}&space;\right)}}$ $={c^{2}{\left(&space;\frac{k^{2}+k}{k+1}&space;\right)}^{2}}$ ∴ বামপক্ষ = ডানপক্ষ (প্রমাণিত)

 Q4. যদি  a : b = b : c  হয়, তবে প্রমান করি যে, (ii) ${\color{Blue}&space;{{a}^{2}}{{b}^{2}}{{c}^{2}}\left(&space;\frac{1}{{{a}^{3}}}+\frac{1}{{{b}^{3}}}+\frac{1}{{{c}^{3}}}&space;\right)={{a}^{3}}+{{b}^{{3}}}+{{c}^{3}}}$ সমাধানঃ a : b = b : c বা, $\frac{a}{b}=\frac{b}{c}=k$ (ধরি) $\therefore&space;b=ck,a=bk=ck\times&space;k=ck^{2}$   বামপক্ষ : ${{a}^{2}}{{b}^{2}}{{c}^{2}}\left(&space;\frac{1}{{{a}^{3}}}+\frac{1}{{{b}^{3}}}+\frac{1}{{{c}^{3}}}&space;\right)$ a ও b এর মান বসিয়ে পাই, $={\left&space;(&space;ck^{2}&space;\right&space;)^{2}}\times&space;{\left&space;(&space;ck&space;\right&space;)^{2}}\times&space;{{c}^{2}}\times&space;\left(&space;\frac{1}{{\left&space;(&space;ck^{2}&space;\right&space;)^{3}}}+\frac{1}{{\left&space;(&space;ck&space;\right&space;)^{3}}}+\frac{1}{{{c}^{3}}}&space;\right)$ $={\left&space;(&space;c^{2}k^{4}&space;\right&space;)}\times&space;{\left&space;(&space;c^{2}k^{2}&space;\right&space;)}\times&space;{{c}^{2}}\times&space;\left[&space;\frac{1}{{\left&space;(&space;c^{3}k^{6}&space;\right&space;)}}+\frac{1}{{\left&space;(&space;c^{3}k^{3}&space;\right&space;)}}+\frac{1}{{{c}^{3}}}&space;\right]$ $=c^{6}k^{6}\times&space;\left&space;(&space;\frac{1+k^{3}+k^{6}}{c^{3}k^{6}}&space;\right&space;)$ $=c^{3}\left&space;(1+k^{3}+k^{6}&space;\right&space;)$   ডানপক্ষ : ${{a}^{3}}+{{b}^{{3}}}+{{c}^{3}}$ a ও b এর মান বসিয়ে পাই, $={{\left&space;(&space;ck^{2}&space;\right&space;)}^{3}}+{{\left&space;(&space;ck&space;\right&space;)}^{{3}}}+{{c}^{3}}$ $={{\left&space;(&space;c^{3}k^{6}&space;\right&space;)}}+{{\left&space;(&space;c^{3}k^{3}&space;\right&space;)}}+{{c}^{3}}$ $=c^{3}\left&space;(1+k^{3}+k^{6}&space;\right&space;)$ ∴ বামপক্ষ = ডানপক্ষ (প্রমাণিত)

 Q4. যদি  a : b = b : c  হয়, তবে প্রমান করি যে, (iii) ${\color{Blue}&space;\frac{abc{{\left(&space;a+b+c&space;\right)}^{3}}}{{{\left(&space;ab+bc+ca&space;\right)}^{3}}}=1}$ সমাধানঃ a : b = b : c বা, $\frac{a}{b}=\frac{b}{c}=k$ (ধরি) $\therefore&space;b=ck,a=bk=ck\times&space;k=ck^{2}$   বামপক্ষ : $\frac{abc{{\left(&space;a+b+c&space;\right)}^{3}}}{{{\left(&space;ab+bc+ca&space;\right)}^{3}}}$ a ও b এর মান বসিয়ে পাই, $=\frac{ck^{2}\times&space;ck\times&space;c\times&space;{{\left(&space;ck^{2}+ck+c&space;\right)}^{3}}}{{{\left(&space;ck^{2}\times&space;ck+ck\times&space;c+c\times&space;ck^{2}&space;\right)}^{3}}}$ $=\frac{c^{3}k^{3}c^{3}{{\left(&space;k^{2}+k+1&space;\right)}^{3}}}{{{\left(&space;c^{2}k^{3}+c^{2}k+c^{2}k^{2}&space;\right)}^{3}}}$ $=\frac{c^{6}k^{3}{{\left(&space;k^{2}+k+1&space;\right)}^{3}}}{{{c^{6}k^{3}\left(&space;k^{2}+1+k&space;\right)}^{3}}}$ = 1 = ডানপক্ষ    ∴ বামপক্ষ = ডানপক্ষ (প্রমাণিত)

 Q5. a, b, c, d  ক্রমিক সমানুপাতী হলে, প্রমাণ করি যে, (i) ${\color{Blue}&space;\left(&space;{{a}^{2}}+{{b}^{2}}+{{c}^{2}}&space;\right)\left(&space;{{b}^{2}}+{{c}^{2}}+{{d}^{2}}&space;\right)={{\left(&space;ab+bc+cd&space;\right)}^{2}}}$ সমাধানঃ a : b = b : c = c : d বা, $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k$ (ধরি) $\therefore&space;c=dk,$ $b=ck=dk\times&space;k=dk^{2}$ ও  $a=bk=dk^{2}\times&space;k=dk^{3}$   বামপক্ষ : $\left(&space;{{a}^{2}}+{{b}^{2}}+{{c}^{2}}&space;\right)\left(&space;{{b}^{2}}+{{c}^{2}}+{{d}^{2}}&space;\right)$ a, b ও c এর মান বসিয়ে পাই, $=\left&space;[\left&space;(&space;dk^{3}&space;\right&space;)^{2}+\left&space;(dk^{2}&space;\right&space;)^{2}+\left&space;(&space;dk&space;\right&space;)^{2}&space;\right&space;]\left&space;[\left&space;(dk^{2}&space;\right&space;)^{2}+\left&space;(&space;dk&space;\right&space;)^{2}+d^{2}&space;\right&space;]$ $=\left&space;[d^{2}k^{6}+d^{2}k^{4}+d^{2}k^{2}&space;\right&space;]\left&space;[d^{2}k^{4}+d^{2}k^{2}+d^{2}&space;\right&space;]$ $=\left&space;[d^{2}k^{2}\left&space;(k^{4}+k^{2}+1&space;\right&space;)\right&space;]\left&space;[d^{2}\left&space;(k^{4}+k^{2}+1&space;\right&space;)&space;\right&space;]$ $=\left&space;[d^{4}k^{2}\left&space;(k^{4}+k^{2}+1&space;\right&space;)^{2}\right&space;]$   ডানপক্ষ : ${{\left(&space;ab+bc+cd&space;\right)}^{2}}$ a, b ও c এর মান বসিয়ে পাই, $={{\left(&space;dk^{3}\times&space;dk^{2}+dk^{2}\times&space;dk+dk\times&space;d&space;\right)}^{2}}$ $={{\left(&space;d^{2}k^{5}+d^{2}k^{3}+d^{2}k&space;\right)}^{2}}$ $=\left&space;[d^{2}k\left&space;(k^{4}+k^{2}+1&space;\right&space;)\right&space;]^{2}$ $=\left&space;[d^{4}k^{2}\left&space;(k^{4}+k^{2}+1&space;\right&space;)^{2}\right&space;]$   ∴ বামপক্ষ = ডানপক্ষ (প্রমাণিত)

 Q5. a, b, c, d  ক্রমিক সমানুপাতী হলে, প্রমাণ করি যে, (ii) ${\color{Blue}&space;{{\left(&space;b-c&space;\right)}^{2}}+{{\left(&space;c-a&space;\right)}^{2}}+{{\left(&space;b-d&space;\right)}^{2}}={{\left(&space;a-d&space;\right)}^{2}}}$ সমাধানঃ a : b = b : c = c : d বা, $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k$ (ধরি) $\therefore&space;c=dk,$ $b=ck=dk\times&space;k=dk^{2}$ ও  $a=bk=dk^{2}\times&space;k=dk^{3}$   বামপক্ষ : ${{\left(&space;b-c&space;\right)}^{2}}+{{\left(&space;c-a&space;\right)}^{2}}+{{\left(&space;b-d&space;\right)}^{2}}$ a, b ও c এর মান বসিয়ে পাই, $={{\left(&space;dk^{2}-dk&space;\right)}^{2}}+{{\left(&space;dk-dk^{3}&space;\right)}^{2}}+{{\left(&space;dk^{2}-d&space;\right)}^{2}}$ $=d^{2}k^{4}-2\times&space;dk^{2}\times&space;dk+d^{2}k^{2}+d^{2}k^{2}-2\times&space;dk\times&space;dk^{3}+d^{2}k^{6}+d^{2}k^{4}-2\times&space;dk^{2}\times&space;d+d^{2}$ $=d^{2}k^{4}-2d^{2}k^{3}+d^{2}k^{2}+d^{2}k^{2}-2&space;d^{2}k^{4}+d^{2}k^{6}+d^{2}k^{4}-2d^{2}k^{2}+d^{2}$ $=2d^{2}k^{4}-2d^{2}k^{3}+2d^{2}k^{2}-2d^{2}k^{4}+d^{2}k^{6}-2d^{2}k^{2}+d^{2}$ $=-2d^{2}k^{3}+d^{2}k^{6}+d^{2}$ $=d^{2}k^{6}-2d^{2}k^{3}+d^{2}$ $=d^{2}\left&space;(k^{6}-2k^{3}+1&space;\right&space;)$ ডানপক্ষ : ${{\left(&space;a-d&space;\right)}^{2}}$ a এর মান বসিয়ে পাই, $={{\left(&space;dk^{3}-d&space;\right)}^{2}}$ $=d^{2}k^{6}-2\times&space;dk^{3}\times&space;d+d^{2}$ $=d^{2}k^{6}-2d^{2}k^{3}+d^{2}$ $=d^{2}\left&space;(k^{6}-2k^{3}+1&space;\right&space;)$   ∴ বামপক্ষ = ডানপক্ষ (প্রমাণিত)

 Q6 (i) যদি  ${\color{Blue}&space;\frac{m}{a}=\frac{n}{b}}$  হয়, তবে দেখাই যে,  ${\color{Blue}&space;\left(&space;{{m}^{2}}+{{n}^{2}}&space;\right)\left(&space;{{a}^{2}}+{{b}^{2}}&space;\right)={{\left(&space;am+bn&space;\right)}^{2}}}$ সমাধানঃ প্রদত্ত, $\frac{m}{a}=\frac{n}{b}=k$ (ধরি) $\therefore&space;m=ak,n=bk$    বামপক্ষ : $\left(&space;{{m}^{2}}+{{n}^{2}}&space;\right)\left(&space;{{a}^{2}}+{{b}^{2}}&space;\right)$ m, ও n এর মান বসিয়ে পাই, $=\left[&space;{{\left&space;(&space;ak&space;\right&space;)}^{2}}+{{\left&space;(&space;bk&space;\right&space;)}^{2}}&space;\right]\times&space;\left(&space;{{a}^{2}}+{{b}^{2}}&space;\right)$ $=\left(&space;{{a}^{2}k^{2}}+{{b}^{2}k^{2}}&space;\right)\times&space;\left(&space;{{a}^{2}}+{{b}^{2}}&space;\right)$ $={{a}^{4}k^{2}}+{a^{2}{b}^{2}k^{2}}+a^{2}{b}^{2}k^{2}+{b}^{4}k^{2}$ $={{a}^{4}k^{2}}+2{a^{2}{b}^{2}k^{2}}+{b}^{4}k^{2}$ $=k^{2}\left&space;({{a}^{4}}+2{a^{2}{b}^{2}}+{b}^{4}&space;\right&space;)$ $=k^{2}\left&space;({{a}^{2}+{b}^{2}}&space;\right&space;)^{2}$   ডানপক্ষ : ${{\left(&space;am+bn&space;\right)}^{2}}$ m, ও n এর মান বসিয়ে পাই, $={{\left(&space;a\times&space;ak+b\times&space;bk&space;\right)}^{2}}$ $=\left&space;(&space;a^{2}k+b^{2}k&space;\right&space;)^{2}$ $=\left&space;(&space;a^{4}k^{2}+2\times&space;a^{2}k\times&space;b^{2}k+b^{4}k^{2}&space;\right&space;)$ $=\left&space;(&space;a^{4}k^{2}+2a^{2}b^{2}k^{2}+b^{4}k^{2}&space;\right&space;)$ $=k^{2}\left&space;({{a}^{4}}+2{a^{2}{b}^{2}}+{b}^{4}&space;\right&space;)$ $=k^{2}\left&space;({{a}^{2}+{b}^{2}}&space;\right&space;)^{2}$   ∴ বামপক্ষ = ডানপক্ষ (প্রমাণিত)

 Q6 (ii) যদি  ${\color{Blue}&space;\frac{a}{b}=\frac{x}{y}}$  হয়, তবে দেখাই যে, ${\color{Blue}&space;\left(&space;a+b&space;\right)\left(&space;{{a}^{2}}+{{b}^{2}}&space;\right){{x}^{3}}=\left(&space;x+y&space;\right)\left(&space;{{x}^{2}}+{{y}^{2}}&space;\right){{a}^{3}}}$ সমাধানঃ প্রদত্ত, $\frac{a}{b}=\frac{x}{y}$ $\frac{a}{x}=\frac{b}{y}=k$ (ধরি) $\therefore&space;a=xk,b=yk$   বামপক্ষ : $\left(&space;a+b&space;\right)\left(&space;{{a}^{2}}+{{b}^{2}}&space;\right){{x}^{3}}$ a, ও b এর মান বসিয়ে পাই, $=\left(&space;xk+yk&space;\right)\times&space;\left[&space;{{\left&space;(&space;xk&space;\right&space;)}^{2}}+{{\left&space;(yk&space;\right&space;)}^{2}}&space;\right]\times&space;x^{3}$ $=\left(&space;xk+yk&space;\right)\times&space;\left[&space;{{&space;x^{2}k^{2}}}+{{y}^{2}k^{2}}&space;\right]\times&space;x^{3}$ $=k\left(&space;x+y&space;\right)\times&space;k^{2}\left[&space;{x^{2}}+{y^{2}}&space;\right]\times&space;x^{3}$ $=x^{3}k^{3}\left(&space;x+y&space;\right)\left(&space;{x^{2}}+{y^{2}}&space;\right)$   ডানপক্ষ : $\left(&space;x+y&space;\right)\left(&space;{{x}^{2}}+{{y}^{2}}&space;\right){{a}^{3}}$ a এর মান বসিয়ে পাই, $=\left(&space;x+y&space;\right)\left(&space;{{x}^{2}}+{{y}^{2}}&space;\right){{\left&space;(&space;xk&space;\right&space;)}^{3}}$ $=x^{3}k^{3}\left(&space;x+y&space;\right)\left(&space;{x^{2}}+{y^{2}}&space;\right)$   ∴ বামপক্ষ = ডানপক্ষ (প্রমাণিত)

 Q6 (iii) যদি  ${\color{Blue}&space;\frac{x}{lm-{{n}^{2}}}=\frac{y}{mn-{{l}^{2}}}=\frac{z}{nl-{{m}^{2}}}}$  হয়, তবে দেখাই যে, ${\color{Blue}&space;lx+my+nz=0}$ সমাধানঃ $\frac{x}{lm-{{n}^{2}}}=\frac{y}{mn-{{l}^{2}}}=\frac{z}{nl-{{m}^{2}}}$ বা, $\frac{x}{lm-{{n}^{2}}}=\frac{y}{mn-{{l}^{2}}}=\frac{z}{nl-{{m}^{2}}}=k$ (ধরি) $\therefore&space;x=k\left&space;({lm-{{n}^{2}}}&space;\right&space;),y=k\left&space;({mn-{{l}^{2}}}&space;\right&space;),z=k\left&space;({nl-{{m}^{2}}}&space;\right&space;)$ বামপক্ষ : $lx+my+nz$ x , y ও z এর মান বসিয়ে পাই, $=l\times&space;k\left&space;({lm-{{n}^{2}}}&space;\right&space;)+m\times&space;k\left&space;({mn-{{l}^{2}}}&space;\right&space;)+n\times&space;k\left&space;({nl-{{m}^{2}}}&space;\right&space;)$ $=k\left&space;[l\times&space;\left&space;({lm-{{n}^{2}}}&space;\right&space;)+m\times&space;\left&space;({mn-{{l}^{2}}}&space;\right&space;)+n\times&space;\left&space;({nl-{{m}^{2}}}&space;\right&space;)&space;\right&space;]$ $=k\left&space;(l^{2}m-{n}^{2}l+m^{2}n-{l}^{2}m+n^{2}l-{m}^{2}n&space;\right&space;)$ $=k\times&space;0$ = 0 = ডানপক্ষ (প্রমাণিত)

 Q6 (iv) ${\color{Blue}&space;\frac{x}{b+c-a}=\frac{y}{c+a-b}=\frac{z}{a+b-c}}$  হলে, দেখাই যে,  ${\color{Blue}&space;\left(&space;b-c&space;\right)x+\left(&space;c-a&space;\right)y+\left(&space;a-b&space;\right)z=0}$ সমাধানঃ $\frac{x}{b+c-a}=\frac{y}{c+a-b}=\frac{z}{a+b-c}$ বা, $\frac{x}{b+c-a}=\frac{y}{c+a-b}=\frac{z}{a+b-c}=k$ (ধরি) $\therefore&space;x=k\left&space;(b+c-a&space;\right&space;),y=k\left&space;(c+a-b&space;\right&space;),z=k\left&space;(a+b-c&space;\right&space;)$   বামপক্ষ : $\left(&space;b-c&space;\right)x+\left(&space;c-a&space;\right)y+\left(&space;a-b&space;\right)z$ x , y ও z এর মান বসিয়ে পাই, $=\left(&space;b-c&space;\right)\times&space;k\left&space;(&space;b+c-a&space;\right&space;)+\left(&space;c-a&space;\right)\times&space;k\left&space;(&space;c+a-b&space;\right&space;)+\left(&space;a-b&space;\right)\times&space;k\left&space;(&space;a+b-c&space;\right&space;)$ $=k\left&space;[\left(&space;b-c&space;\right)\times&space;\left&space;(&space;b+c-a&space;\right&space;)+\left(&space;c-a&space;\right)\times&space;\left&space;(&space;c+a-b&space;\right&space;)+\left(&space;a-b&space;\right)\times&space;\left&space;(&space;a+b-c&space;\right&space;)&space;\right&space;]$ $=k\left&space;[b^{2}+bc-ab-bc-c^{2}+ac+c^{2}+ac-bc-ac-a^{2}+ab+a^{2}+ab-ac-ab-b^{2}+bc&space;\right&space;]$ $=k\times&space;0$ = 0 = ডানপক্ষ (প্রমাণিত)

 Q6 (v) ${\color{Blue}&space;\frac{x}{y}=\frac{a+2}{a-2}}$  হলে দেখাই যে, ${\color{Blue}&space;\frac{{{x}^{2}}-{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}}=\frac{4a}{{{a}^{2}}+4}}$ সমাধানঃ $\frac{x}{y}=\frac{a+2}{a-2}$  উভয়পক্ষে বর্গ করে পাই, বা, $\frac{x^{2}}{y^{2}}=\frac{\left&space;(a+2&space;\right&space;)^{2}}{\left&space;(a-2&space;\right&space;)^{2}}$   বা, $\frac{x^{2}}{y^{2}}=\frac{a^{2}+4a+4}{a^{2}-4a+4}$   বা, $\frac{x^{2}+y^{2}}{x^{2}-y^{2}}=\frac{\left&space;(a^{2}+4a+4&space;\right&space;)+\left&space;(a^{2}-4a+4&space;\right&space;)}{\left&space;(a^{2}+4a+4&space;\right&space;)-\left&space;(a^{2}-4a+4&space;\right&space;)}$ [ যোগ ভাগ প্রক্রিয়া করে পাই ]   বা, $\frac{x^{2}+y^{2}}{x^{2}-y^{2}}=\frac{a^{2}+4a+4&space;+a^{2}-4a+4&space;}{a^{2}+4a+4&space;-a^{2}+4a-4}$ বা, $\frac{x^{2}+y^{2}}{x^{2}-y^{2}}=\frac{2\left&space;(a^{2}+4&space;\right&space;)}{8a}$ বা, $\frac{x^{2}+y^{2}}{x^{2}-y^{2}}=\frac{\left&space;(a^{2}+4&space;\right&space;)}{4a}$ $\therefore&space;\frac{{{x}^{2}}-{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}}=\frac{4a}{{{a}^{2}}+4}$ (প্রমাণিত)

 Q6 (vi) ${\color{Blue}&space;x=\frac{8ab}{a+b}}$  হলে,  ${\color{Blue}&space;\left(&space;\frac{x+4a}{x-4a}+\frac{x+4b}{x-4b}&space;\right)}$  -এর মান হিসাব করে লিখি। সমাধানঃ প্রদত্ত, $x=\frac{8ab}{a+b}$  বা, $\frac{x}{4a}=\frac{2b}{a+b}$  বা, $\frac{x+4a}{x-4a}=\frac{2b+a+b}{2b-a-b}$  বা, $\frac{x+4a}{x-4a}=\frac{3b+a}{b-a}$    আবার, $x=\frac{8ab}{a+b}$   বা, $\frac{x}{4b}=\frac{2a}{a+b}$  বা, $\frac{x+4b}{x-4b}=\frac{2a+a+b}{2a-a-b}$  বা, $\frac{x+4b}{x-4b}=\frac{3a+b}{a-b}$    ∴ $\left(&space;\frac{x+4a}{x-4a}+\frac{x+4b}{x-4b}&space;\right)$ -এর মান $=\frac{3b+a}{b-a}+\frac{3a+b}{a-b}$ $=\frac{3b+a}{b-a}+\frac{3a+b}{-\left&space;(b-a&space;\right&space;)}$ $=\frac{3b+a}{b-a}-\frac{3a+b}{b-a}$ $=\frac{3b+a-3a-b}{b-a}$ $=\frac{2b-2a}{b-a}$ $=\frac{2\left&space;(b-a&space;\right&space;)}{\left&space;(b-a&space;\right&space;)}$  = 2 (উত্তর)

Koshe dekhi 5.3 class 10

 Q7 (i) ${\color{Blue}&space;\frac{a}{3}=\frac{b}{4}=\frac{c}{7}}$  হলে দেখাই যে, ${\color{Blue}&space;\frac{a+b+c}{c}=2}$ সমাধানঃ প্রদত্ত, $\frac{a}{3}=\frac{b}{4}=\frac{c}{7}=k$ $\therefore&space;a=3k,b=4k,c=7k$  (ধরি)   বামপক্ষ : $\frac{a+b+c}{c}$ $=\frac{3k+4k+7k}{7k}$ $=\frac{14k}{7k}$ = 2 (প্রমাণিত)

Koshe dekhi 5.3 class 10

 Q7 (ii) ${\color{Blue}&space;\frac{a}{q-r}=\frac{b}{r-p}=\frac{c}{p-q}}$  হলে দেখাই যে, ${\color{Blue}&space;a+b+c=0=pa+qb+rc}$ সমাধানঃ প্রদত্ত, $\frac{a}{q-r}=\frac{b}{r-p}=\frac{c}{p-q}=k$ (ধরি) $\therefore&space;a=k\left&space;(&space;q-r&space;\right&space;),b=k\left&space;(&space;r-p&space;\right&space;),c=k\left&space;(&space;p-q&space;\right&space;)$ $\therefore&space;a+b+c$ $=k\left&space;(&space;q-r&space;\right&space;)+k\left&space;(&space;r-p&space;\right&space;)+k\left&space;(&space;p-q&space;\right&space;)$ $=kq-kr+kr-pk+pk-qk$ = 0   আবার, $pa+qb+rc$ $=p\times&space;k\left&space;(&space;q-r&space;\right&space;)+q\times&space;k\left&space;(&space;r-p&space;\right&space;)+r\times&space;k\left&space;(&space;p-q&space;\right&space;)$ $=pqk-prk+qrk-pqk+prk-qrk$  = 0 ${\color{DarkGreen}&space;\therefore&space;a+b+c=0=pa+qb+rc}$ (প্রমাণিত)

Koshe dekhi 5.3 class 10

 Q7 (iii) ${\color{Blue}&space;\frac{ax+by}{a}=\frac{bx-ay}{b}}$  হলে দেখাই যে, প্রতিটি অনুপাত  x -এর সমান। সমাধানঃ $\frac{ax+by}{a}=\frac{bx-ay}{b}$ প্রথম অনুপাতের উভয় পদকে a দ্বারা এবং দ্বিতীয় অনুপাতের উভয় পদকে b দ্বারা গুন্ করে পাই, $\frac{a^{2}x+aby}{a^{2}}=\frac{b^{2}x-aby}{b^{2}}$ সংযোজন প্রক্রিয়া করে পাই, $=\frac{a^{2}x+aby+b^{2}x-aby}{a^{2}+b^{2}}$ $=\frac{x\left&space;(a^{2}+b^{2}&space;\right&space;)}{a^{2}+b^{2}}$ = x  ${\color{DarkGreen}&space;\therefore&space;\frac{ax+by}{a}=\frac{bx-ay}{b}=x}$ (প্রমাণিত)

Koshe dekhi 5.3 class 10

 Q8 (i) যদি  ${\color{Blue}&space;\frac{a+b}{b+c}=\frac{c+d}{d+a}}$  হয়, তবে প্রমান করি যে,  c = a  অথবা  a + b + c + d = 0 সমাধানঃ $\frac{a+b}{b+c}=\frac{c+d}{d+a}$  বা, $\left&space;(&space;a+b&space;\right&space;)\left&space;(&space;d+a&space;\right&space;)=\left&space;(&space;b+c&space;\right&space;)\left&space;(&space;c+d&space;\right&space;)$ বা, $ad+a^{2}+bd+ab=bc+bd+c^{2}+cd$ বা, $ad+a^{2}+bd+ab-bc-bd-c^{2}-cd=0$ বা, $ad-cd+ab-bc+a^{2}-c^{2}=0$ বা, $d\left&space;(&space;a-c&space;\right&space;)+b\left&space;(&space;a-c&space;\right&space;)+\left&space;(&space;a+c&space;\right&space;)\left&space;(&space;a-c&space;\right&space;)=0$ বা, $\left&space;(&space;a-c&space;\right&space;)\left&space;(&space;d+b+a+c&space;\right&space;)=0$ দুই বা ততোধিক রাশির গুনফল শূন্য হলে তারা প্রত্যেকে পৃথক পৃথক ভাবে শূন্য হবে।  সুতরাং, a – c = 0 a = c অথবা  a + b + c + d = 0 ∴ c = a  অথবা  a + b + c + d = 0 (প্রমাণিত)

Koshe dekhi 5.3 class 10

 Q8 (ii) যদি  ${\color{Blue}&space;\frac{x}{b+c}=\frac{y}{c+a}=\frac{z}{a+b}}$  হয়, দেখাই যে, ${\color{Blue}&space;\frac{a}{y+z-x}=\frac{b}{z+x-y}=\frac{c}{x+y-z}}$ সমাধানঃ প্রদত্ত,  $\frac{x}{b+c}=\frac{y}{c+a}=\frac{z}{a+b}$  ধরি,  $\frac{x}{b+c}=\frac{y}{c+a}=\frac{z}{a+b}=k$  $\therefore&space;x=k\left&space;(&space;b+c&space;\right&space;),y=k\left&space;(&space;c+a&space;\right&space;),z=k\left&space;(&space;a+b&space;\right&space;)$ প্রথমপক্ষ : $=\frac{a}{y+z-x}$ $=\frac{a}{k\left&space;(&space;c+a&space;\right&space;)+k\left&space;(&space;a+b&space;\right&space;)-k\left&space;(&space;b+c&space;\right&space;)}$ $=\frac{a}{k\left&space;(&space;c+a+a+b-b-c&space;\right&space;)}$ $=\frac{a}{2ak}$ $=\frac{1}{2k}$ দ্বিতীয়পক্ষ :  $=\frac{b}{z+x-y}$ $=\frac{b}{k\left&space;(&space;a+b&space;\right&space;)+k\left&space;(&space;b+c&space;\right&space;)-k\left&space;(&space;c+a&space;\right&space;)}$ $=\frac{b}{k\left&space;(&space;a+b+b+c-c-a&space;\right&space;)}$ $=\frac{b}{2bk}$ $=\frac{1}{2k}$ তৃতীয়পক্ষ : $=\frac{c}{x+y-z}$ $=\frac{c}{k\left&space;(&space;b+c+c+a-a-b&space;\right&space;)}$ $=\frac{c}{2ck}$ $=\frac{1}{2k}$ ${\color{DarkGreen}&space;\therefore&space;\frac{a}{y+z-x}=\frac{b}{z+x-y}=\frac{c}{x+y-z}}$ (প্রমাণিত)

Koshe dekhi 5.3 class 10

 Q8 (iii) ${\color{Blue}&space;\frac{x+y}{3a-b}=\frac{y+z}{3b-c}=\frac{z+x}{3c-a}}$  হলে, দেখাই যে, ${\color{Blue}&space;\frac{x+y+z}{a+b+c}=\frac{ax+by+cz}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$ সমাধানঃ প্রদত্ত, $\frac{x+y}{3a-b}=\frac{y+z}{3b-c}=\frac{z+x}{3c-a}$   ধরি, $\frac{x+y}{3a-b}=\frac{y+z}{3b-c}=\frac{z+x}{3c-a}=k$ $\therefore&space;x+y=k\left&space;(3a-b&space;\right&space;)$ ……….(i) $y+z=k\left&space;(3b-c&space;\right&space;)$ ……….(ii) $z+x=k\left&space;(3c-a&space;\right&space;)$ ………(iii)   (i) + (ii) + (iii) সমীকরণ যোগ করে পাই, $x+y+y+z+z+x=k\left&space;(&space;3a-b&space;\right&space;)+k\left&space;(3b-c&space;\right&space;)+k\left&space;(&space;3c-a&space;\right&space;)$ বা, $\left&space;(2x+2y+2z&space;\right&space;)=k\left&space;(&space;3a-b+3b-c+3c-a&space;\right&space;)$ বা, $2\left&space;(x+y+z&space;\right&space;)=k\left&space;(&space;2a+2b+2c&space;\right&space;)$ বা, $2\left&space;(x+y+z&space;\right&space;)=2k\left&space;(&space;a+b+c&space;\right&space;)$ বা, $\left&space;(x+y+z&space;\right&space;)=k\left&space;(&space;a+b+c&space;\right&space;)$  ………(iv) $\therefore&space;\frac{x+y+z}{a+b+c}=k$    (iv) নং সমীকরণ থেকে (i) নং সমীকরণ বিয়োগ করে পাই, $\left&space;(x+y+z&space;\right&space;)-\left&space;(&space;x+y&space;\right&space;)=k\left&space;(&space;a+b+c&space;\right&space;)-k\left&space;(&space;3a-b&space;\right&space;)$ বা, $x+y+z-x-y&space;=k\left&space;(&space;a+b+c-3a+b&space;\right&space;)$ বা, $z=k\left&space;(2b+c-2a&space;\right&space;)$   (iv) নং সমীকরণ থেকে (ii) নং সমীকরণ বিয়োগ করে পাই, বা, $\left&space;(x+y+z&space;\right&space;)-\left&space;(&space;y+z&space;\right&space;)=k\left&space;(&space;a+b+c&space;\right&space;)-k\left&space;(&space;3b-c&space;\right&space;)$ বা, $\left&space;(x+y+z-y-z&space;\right&space;)=k\left&space;(&space;a+b+c-3b+c&space;\right&space;)$ বা, $x=k\left&space;(2c+a-2b&space;\right&space;)$   (iv) নং সমীকরণ থেকে (iii) নং সমীকরণ বিয়োগ করে পাই, $\left&space;(x+y+z&space;\right&space;)-\left&space;(&space;z+x&space;\right&space;)=k\left&space;(&space;a+b+c&space;\right&space;)-k\left&space;(&space;3c-a&space;\right&space;)$ বা, $x+y+z-z-x&space;=k\left&space;(&space;a+b+c-3c+a&space;\right&space;)$ বা, $y=k\left&space;(2a+b-2c&space;\right&space;)$   আবার, $\frac{ax+by+cz}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}$ $=\frac{a\times&space;k\left&space;(&space;2c+a-2b&space;\right&space;)+b\times&space;\left&space;(&space;2a+b-2c&space;\right&space;)+c\times&space;\left&space;(&space;2b+c-2a&space;\right&space;)}{a^{2}+b^{2}+c^{2}}$ $=\frac{k\left&space;(&space;2ac+a^{2}-2ab+2ab+b^{2}-2bc+2bc+c^{2}-2ac&space;\right&space;)}{a^{2}+b^{2}+c^{2}}$ $=\frac{k\left&space;(&space;a^{2}+b^{2}+c^{2}&space;\right&space;)}{\left&space;(a^{2}+b^{2}+c^{2}&space;\right&space;)}$ = k   ${\color{DarkGreen}&space;\therefore&space;\frac{x+y+z}{a+b+c}=\frac{ax+by+cz}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}$ (প্রমাণিত)

Koshe dekhi 5.3 class 10

 Q8 (iv) ${\color{Blue}&space;\frac{x}{a}=\frac{y}{b}=\frac{z}{c}}$  হলে, দেখাই যে, ${\color{Blue}&space;\frac{{{x}^{2}}-yz}{{{a}^{2}}-bc}=\frac{{{y}^{2}}-zx}{{{b}^{2}}-ca}=\frac{{{z}^{2}}-xy}{{{c}^{2}}-ab}}$ সমাধানঃ প্রদত্ত, $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$  ধরি, $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k$  $\therefore&space;x=ak,y=bk,z=ck$ প্রথমপক্ষ : $\frac{{{x}^{2}}-yz}{{{a}^{2}}-bc}$ $=\frac{\left&space;(&space;ak&space;\right&space;)^{2}-bk\times&space;ck}{a^{2}-bc}$ $=\frac{a^{2}k^{2}-bck^{2}}{a^{2}-bc}$ $=\frac{k^{2}\left&space;(a^{2}-bc&space;\right&space;)}{\left&space;(a^{2}-bc&space;\right&space;)}$ $=k^{2}$ দ্বিতীয়পক্ষ :  $\frac{{{y}^{2}}-zx}{{{b}^{2}}-ca}$ $=\frac{\left&space;(&space;bk&space;\right&space;)^{2}-ck\times&space;ak}{b^{2}-ca}$ $=\frac{b^{2}k^{2}-ack^{2}}{b^{2}-ac}$ $=\frac{k^{2}\left&space;(b^{2}-ac&space;\right&space;)}{\left&space;(b^{2}-ac&space;\right&space;)}$ $=k^{2}$ তৃতীয়পক্ষ : $\frac{{{z}^{2}}-xy}{{{c}^{2}}-ab}$ $=\frac{\left&space;(&space;ck&space;\right&space;)^{2}-ak\times&space;bk}{c^{2}-ab}$ $=\frac{c^{2}k^{2}-abk^{2}}{c^{2}-ab}$ $=\frac{k^{2}\left&space;(c^{2}-ab&space;\right&space;)}{\left&space;(c^{2}-ab&space;\right&space;)}$ $=k^{2}$ ${\color{DarkGreen}&space;\therefore&space;\frac{{{x}^{2}}-yz}{{{a}^{2}}-bc}=\frac{{{y}^{2}}-zx}{{{b}^{2}}-ca}=\frac{{{z}^{2}}-xy}{{{c}^{2}}-ab}}$ (প্রমাণিত)
 Q9 (i) যদি  ${\color{Blue}&space;\frac{3x+4y}{3u+4v}=\frac{3x-4y}{3u-4v}}$  হয়, তবে দেখাই যে, ${\color{Blue}&space;\frac{x}{y}=\frac{u}{v}}$ সমাধানঃ  প্রদত্ত, $\frac{3x+4y}{3u+4v}=\frac{3x-4y}{3u-4v}$    উভয়পক্ষে যোগ ভাগ প্রক্রিয়া করে পাই, বা, $\frac{3x+4y+3x-4y}{3x+4y-3x+4y}=\frac{3u+4v+3u-4v}{3u+4v-3u+4v}$   বা, $\frac{6x}{8y}=\frac{6u}{8v}$ ${\color{DarkGreen}&space;\therefore&space;\frac{x}{y}=\frac{u}{v}}$ (প্রমাণিত)

 Q9 (ii) ${\color{Blue}&space;\left(&space;a+b+c+d&space;\right):\left(&space;a+b-c-d&space;\right)=\left(&space;a-b+c-d&space;\right):\left(&space;a-b-c+d&space;\right)}$ হলে, প্রমান করি যে, a : b = c : d. সমাধানঃ প্রদত্ত, $\left(&space;a+b+c+d&space;\right):\left(&space;a+b-c-d&space;\right)=\left(&space;a-b+c-d&space;\right):\left(&space;a-b-c+d&space;\right)$ বা, $\frac{a+b+c+d}{&space;a+b-c-d}=\frac{a-b+c-d}{&space;a-b-c+d}$ বা, $\frac{a+b+c+d+a+b-c-d}{a+b+c+d-a-b+c+d}=\frac{a-b+c-d+a-b-c+d}{a-b+c-d-a+b+c-d}$ বা, $\frac{2a+2b}{2c+2d}=\frac{2a-2b}{2c-2d}$ বা, $\frac{a+b}{c+d}=\frac{a-b}{c-d}$ বা, $\frac{a+b}{a-b}=\frac{c+d}{c-d}$ বা, $\frac{a+b+a-b}{a+b-a+b}=\frac{c+d+c-d}{c+d-c+d}$ বা, $\frac{2a}{2b}=\frac{2c}{2d}$ বা, $\frac{a}{b}=\frac{c}{d}$ ${\color{DarkGreen}&space;\therefore&space;a:b=c:d}$ (প্রমাণিত)

 Q10 (i) ${\color{Blue}&space;\frac{{{a}^{2}}}{b+c}=\frac{{{b}^{2}}}{c+a}=\frac{{{c}^{2}}}{a+b}=1}$  হলে, দেখাই যে, ${\color{Blue}&space;\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=1}$ সমাধানঃ প্রদত্ত, $\frac{{{a}^{2}}}{b+c}=\frac{{{b}^{2}}}{c+a}=\frac{{{c}^{2}}}{a+b}=1$ $\frac{{{a}^{2}}}{b+c}=1$ বা, ${a}^{2}=b+c$ বা, $a+{a}^{2}=a+b+c$ বা, $a\left&space;(&space;1+a&space;\right&space;)=a+b+c$ বা, $\frac{a}{a+b+c}=\frac{1}{1+a}$  ………(i) আবার, $\frac{{{b}^{2}}}{c+a}=1$ বা, ${b}^{2}=c+a$ বা, $b+{b}^{2}=c+a+b$ বা, $b\left&space;(&space;1+b&space;\right&space;)=a+b+c$ বা, $\frac{b}{a+b+c}=\frac{1}{1+b}$ ………(ii) $\frac{{{c}^{2}}}{a+b}=1$ বা, ${c}^{2}=a+b$ বা, $c+{c}^{2}=a+b+c$ বা, $c\left&space;(&space;1+c&space;\right&space;)=a+b+c$ বা, $\frac{c}{a+b+c}=\frac{1}{1+c}$ ………(iii)   বামপক্ষ : $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}$ $=\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}$ $=\frac{a+b+c}{a+b+c}$  = 1   = ডানপক্ষ (প্রমাণিত )

 Q10 (ii) ${\color{Blue}&space;{{x}^{2}}:\left(&space;by+cz&space;\right)={{y}^{2}}:\left(&space;cz+ax&space;\right)={{z}^{2}}:\left(&space;ax+by&space;\right)=1}$ হলে, দেখাই যে, ${\color{Blue}&space;\frac{a}{a+x}+\frac{b}{b+y}+\frac{c}{c+z}=1}$ সমাধানঃ প্রদত্ত, ${{x}^{2}}:\left(&space;by+cz&space;\right)={{y}^{2}}:\left(&space;cz+ax&space;\right)={{z}^{2}}:\left(&space;ax+by&space;\right)=1$ বা, $\frac{{x}^{2}}{by+cz}=\frac{{y}^{2}}{cz+ax}=\frac{{z}^{2}}{ax+by}=1$ $\therefore&space;{x}^{2}=by+cz,{y}^{2}=cz+ax,{z}^{2}=ax+by$ এখন, বামপক্ষ : $\frac{a}{a+x}+\frac{b}{b+y}+\frac{c}{c+z}$ $=\frac{ax}{ax+x^{2}}+\frac{by}{by+y^{2}}+\frac{cz}{cz+z^{2}}$ $=\frac{ax}{ax+by+cz}+\frac{by}{by+cz+ax}+\frac{cz}{cz+ax+by}$ ${\color{Orchid}\left&space;[&space;\because&space;{x}^{2}=by+cz,{y}^{2}=cz+ax,{z}^{2}=ax+by&space;\right&space;]}$ $=\frac{ax+by+cz}{ax+by+cz}$  = 1  = ডানপক্ষ (প্রমাণিত )

 Q11 (i)  ${\color{Blue}&space;\frac{x}{xa+yb+zc}=\frac{y}{ya+zb+xc}=\frac{z}{za+xb+yc}}$   এবং  ${\color{Blue}&space;x+y+z\neq&space;0}$  হলে, দেখাই যে, প্রতিটি অনুপাত  ${\color{Blue}&space;\frac{1}{a+b+c}}$  -এর সমান। সমাধানঃ প্রদত্ত, $\frac{x}{xa+yb+zc}=\frac{y}{ya+zb+xc}=\frac{z}{za+xb+yc}$  সংযোজন প্রক্রিয়া করে পাই, $=\frac{x+y+z}{xa+yb+zc+ya+zb+xc+za+xb+yc}$   $=\frac{x+y+z}{xa+xb+xc+ya+yb+yc+za+zb+zc}$   $=\frac{x+y+z}{x\left&space;(a+b+c&space;\right&space;)+y\left&space;(a+b+c&space;\right&space;)+z\left&space;(a+b+c&space;\right&space;)}$   $=\frac{\left&space;(x+y+z&space;\right&space;)}{\left&space;(a+b+c&space;\right&space;)\left&space;(&space;x+y+z&space;\right&space;)}$ $=\frac{1}{\left&space;(a+b+c&space;\right&space;)}$ ${\color{DarkGreen}&space;\therefore&space;\frac{x}{xa+yb+zc}=\frac{y}{ya+zb+xc}=\frac{z}{za+xb+yc}=\frac{1}{a+b+c}}$  (প্রমাণিত )

 Q11 (ii)  ${\color{Blue}&space;\frac{{{x}^{2}}-yz}{a}=\frac{{{y}^{2}}-zx}{b}=\frac{{{z}^{2}}-xy}{c}}$  হলে, প্রমাণ করি যে,  ${\color{Blue}&space;\left(&space;a+b+c&space;\right)\left(&space;x+y+z&space;\right)=\left(&space;ax+by+cz&space;\right)}$ সমাধানঃ  প্রদত্ত,     $\frac{{{x}^{2}}-yz}{a}=\frac{{{y}^{2}}-zx}{b}=\frac{{{z}^{2}}-xy}{c}$   সংযোজন প্রক্রিয়া করে পাই,   $=\frac{{{x}^{2}}-yz+{{y}^{2}}-zx+{{z}^{2}-xy}}{a+b+c}$ আবার,     $\frac{{{x}^{2}}-yz}{a}=\frac{{{y}^{2}}-zx}{b}=\frac{{{z}^{2}}-xy}{c}$  বা,  $\frac{{x\left&space;({x}^{2}-yz&space;\right&space;)}}{ax}=\frac{{y\left&space;({y}^{2}-zx&space;\right&space;)}}{by}=\frac{{z\left&space;({z}^{2}-xy&space;\right&space;)}}{cz}$   বা, $\frac{{{x}^{3}}-xyz}{ax}=\frac{{{y}^{3}}-yzx}{by}=\frac{{{z}^{3}}-xyz}{cz}$ সংযোজন প্রক্রিয়া করে পাই,  $=\frac{{{x}^{3}+{y}^{3}+{z}^{3}}-3xyz}{ax+by+cz}$  $=\frac{\left&space;(&space;x+y+z&space;\right&space;)\left&space;(&space;x^{2}+y^{2}+z^{2}-xy-yz-zx&space;\right&space;)}{ax+by+cz}$ $\left(&space;a+b+c&space;\right)\left(&space;x+y+z&space;\right)=\left(&space;ax+by+cz&space;\right)$ ${\color{DarkGreen}&space;\therefore&space;\left(&space;a+b+c&space;\right)\left(&space;x+y+z&space;\right)=\left(&space;ax+by+cz&space;\right)}$ (প্রমাণিত )

 Q11 (iii)  ${\color{Blue}&space;\frac{a}{y+z}=\frac{b}{z+x}=\frac{c}{x+y}}$  হলে, প্রমাণ করি যে, ${\color{Blue}&space;\frac{a\left(&space;b-c&space;\right)}{{{y}^{2}}-{{z}^{2}}}=\frac{b\left(&space;c-a&space;\right)}{{{z}^{2}}-{{x}^{2}}}=\frac{c\left(&space;a-b&space;\right)}{{{x}^{2}}-{{y}^{2}}}}$ সমাধানঃ প্রদত্ত, $\frac{a}{y+z}=\frac{b}{z+x}=\frac{c}{x+y}$ ধরি, $\frac{a}{y+z}=\frac{b}{z+x}=\frac{c}{x+y}=k$ $\therefore&space;a=k\left&space;(y+z&space;\right&space;),b=k\left&space;(z+x&space;\right&space;),c=k\left&space;(x+y&space;\right&space;)$ প্রথমপক্ষ : $\frac{a\left(&space;b-c&space;\right)}{{{y}^{2}}-{{z}^{2}}}$ a, b ও c এর মান বসিয়ে পাই, $=\frac{k\left&space;(&space;y+z&space;\right&space;)\times&space;\left&space;[k\left(z+x&space;\right)-k\left&space;(&space;x+y&space;\right&space;)&space;\right&space;]}{{{y}^{2}}-{{z}^{2}}}$ $=\frac{k\left&space;(&space;y+z&space;\right&space;)\times&space;k\left&space;[\left(z+x-x-y&space;\right)&space;\right&space;]}{{{y}^{2}}-{{z}^{2}}}$ $=\frac{k^{2}\left&space;(&space;y+z&space;\right&space;)\left&space;(&space;z-y&space;\right&space;)}{\left&space;(&space;y+z&space;\right&space;)\left&space;(&space;y-z&space;\right&space;)}$ $=\frac{k^{2}\left&space;(&space;y+z&space;\right&space;)\left&space;[-\left&space;(&space;y-z&space;\right&space;)&space;\right&space;]}{\left&space;(&space;y+z&space;\right&space;)\left&space;(&space;y-z&space;\right&space;)}$ $=\frac{-k^{2}\left&space;(&space;y+z&space;\right&space;)\left&space;(&space;y-z&space;\right&space;)&space;}{\left&space;(&space;y+z&space;\right&space;)\left&space;(&space;y-z&space;\right&space;)}$ $=-k^{2}$ দ্বিতীয়পক্ষ : $\frac{b\left(&space;c-a&space;\right)}{{{z}^{2}}-{{x}^{2}}}$ a, b ও c এর মান বসিয়ে পাই, $=\frac{k\left&space;(&space;z+x&space;\right&space;)\times&space;\left&space;[k\left(x+y&space;\right)-k\left&space;(&space;y+z&space;\right&space;)&space;\right&space;]}{{{z}^{2}}-{{x}^{2}}}$ $=\frac{k\left&space;(&space;z+x&space;\right&space;)\times&space;k\left&space;[\left(x+y-y-z&space;\right)&space;\right&space;]}{{{z}^{2}}-{{x}^{2}}}$ $=\frac{k^{2}\left&space;(&space;z+x&space;\right&space;)\left&space;(&space;x-z&space;\right&space;)}{\left&space;(&space;z+x&space;\right&space;)\left&space;(&space;z-x&space;\right&space;)}$ $=\frac{k^{2}\left&space;(&space;z+x&space;\right&space;)\left&space;[-\left&space;(&space;z-x&space;\right&space;)&space;\right&space;]}{\left&space;(&space;z+x&space;\right&space;)\left&space;(&space;z-x&space;\right&space;)}$ $=\frac{-k^{2}\left&space;(&space;z+x&space;\right&space;)\left&space;(&space;z-x&space;\right&space;)}{\left&space;(&space;z+x&space;\right&space;)\left&space;(&space;z-x&space;\right&space;)}$ $=-k^{2}$ তৃতীয়পক্ষ : $\frac{c\left(&space;a-b&space;\right)}{{{x}^{2}}-{{y}^{2}}}$ a, b ও c এর মান বসিয়ে পাই, $=\frac{k\left&space;(&space;x+y&space;\right&space;)\times&space;\left&space;[k\left(y+z&space;\right)-k\left&space;(&space;z+x&space;\right&space;)&space;\right&space;]}{{{x}^{2}}-{{y}^{2}}}$ $=\frac{k\left&space;(&space;x+y&space;\right&space;)\times&space;k\left&space;[\left(y+z-z-x&space;\right)&space;\right&space;]}{{{x}^{2}}-{{y}^{2}}}$ $=\frac{k^{2}\left&space;(&space;x+y&space;\right&space;)\left&space;(&space;y-x&space;\right&space;)}{\left&space;(&space;x+y&space;\right&space;)\left&space;(&space;x-y&space;\right&space;)}$ $=\frac{k^{2}\left&space;(&space;x+y&space;\right&space;)\left&space;[-\left&space;(&space;x-y&space;\right&space;)&space;\right&space;]}{\left&space;(&space;x+y&space;\right&space;)\left&space;(&space;x-y&space;\right&space;)}$ $=\frac{-k^{2}\left&space;(&space;x+y&space;\right&space;)\left&space;(&space;x-y&space;\right&space;)}{\left&space;(&space;x+y&space;\right&space;)\left&space;(&space;x-y&space;\right&space;)}$ $=-k^{2}$ ${\color{DarkGreen}&space;\therefore&space;\frac{a\left(&space;b-c&space;\right)}{{{y}^{2}}-{{z}^{2}}}=\frac{b\left(&space;c-a&space;\right)}{{{z}^{2}}-{{x}^{2}}}=\frac{c\left(&space;a-b&space;\right)}{{{x}^{2}}-{{y}^{2}}}}$ (প্রমাণিত )
 Q12. অতিসংক্ষিপ্ত উত্তরধর্মী প্রশ্ন (V.S.A.) (A) বহুবিকল্পী প্রশ্ন (M.C.Q) : (i) 3, 4 এবং 6 -এর চতুর্থ সমানুপাতী (a) 8 (b) 10 (c) 12 (d) 24 সমাধানঃ ধরি, চতুর্থ সমানুপাতীটি হল x  $\therefore&space;3:4::6:x$ বা, $\frac{3}{4}=\frac{6}{x}$ বা, $3x=24$ বা, $x=\frac{24}{3}$ $\therefore&space;x=8$ উত্তরঃ  (a) 8

 Q12.(A) (ii) 8 এবং 12 -এর তৃতীয় সমানুপাতী (a) 12 (b) 16 (c) 18 (d) 20 সমাধানঃ ধরি, তৃতীয় সমানুপাতীটি হল x  $\therefore&space;8:12::12:x$ বা, $\frac{8}{12}=\frac{12}{x}$ বা, $8x=12\times&space;12$ বা, $x=\frac{12\times&space;12}{8}$ $\therefore&space;x=18$ উত্তরঃ (c) 18

 Q12.(A) (iii) 16 এবং 25 -এর মধ্য সমানুপাতী (a) 400 (b) 100 (c) 20 (d) 40 সমাধানঃ ধরি, মধ্য সমানুপাতীটি হল x  $\therefore&space;16:x::x:25$ বা, $\frac{16}{x}=\frac{x}{25}$ বা, $x^{2}=16\times&space;25$ বা, $x=\sqrt{16\times&space;25}$ $\therefore&space;x=20$ উত্তরঃ (c) 20

 Q12.(A) (iv) a একটি ধনাত্মক সংখ্যা এবং  ${\color{Blue}&space;a:\frac{27}{64}=\frac{3}{4}:a}$ হলে, a -এর মান (a) ${\color{Blue}&space;\frac{81}{256}}$ (b) 9 (c) ${\color{Blue}&space;\frac{9}{16}}$ (d) ${\color{Blue}&space;\frac{16}{9}}$ সমাধানঃ $a:\frac{27}{64}=\frac{3}{4}:a$ বা, $\frac{a}{\frac{27}{64}}=\frac{\frac{3}{4}}{a}$ বা, $a^{2}=\frac{27}{64}\times&space;\frac{3}{4}$ বা, $a=\pm&space;\sqrt{\frac{27}{64}\times&space;\frac{3}{4}}$ $\therefore&space;a=\frac{9}{16}$  [ ${\color{Blue}&space;\because&space;}$ a একটি ধনাত্মক সংখ্যা ] উত্তরঃ (c) ${\color{DarkGreen}&space;\frac{9}{16}}$

 Q12.(A) (v)  2a = 3b = 4c হলে,  a : b : c হবে (a) 3 : 4 : 6 (b) 4 : 3 : 6 (c) 3 : 6 : 4 (d) 6 : 4 : 3 সমাধানঃ প্রদত্ত, 2a = 3b = 4c এখন, 2a = 3b বা, $\frac{a}{b}=\frac{3}{2}$ বা, $\frac{a}{b}=\frac{3}{2}=\frac{3\times&space;2}{2\times&space;2}=\frac{6}{4}$ আবার, 3b = 4c বা, $\frac{b}{c}=\frac{4}{3}$ $\therefore&space;a:b:c=6:4:3$ উত্তরঃ (d) 6 : 4 : 3
 Q12. (B) নীচের বিবৃতিগুলি সত্য না মিথ্যা লিখি : (i) ${\color{Blue}&space;ab:{{c}^{2}},\&space;bc:{{a}^{2}}}$  এবং  ${\color{Blue}&space;ca:{{b}^{2}}}$  -এর যৌগিক অনুপাত 1 : 1 সমাধানঃ $ab:{{c}^{2}},\&space;bc:{{a}^{2}}$  এবং  $ca:{{b}^{2}}$ -এর যৌগিক অনুপাত $=ab\times&space;bc\times&space;ca:c^{2}\times&space;a^{2}\times&space;b^{2}$ $=c^{2}a^{2}b^{2}:c^{2}a^{2}b^{2}$ = 1 : 1 উত্তরঃ বিবৃতিটি সত্য

 Q12. (B) (ii) ${\color{Blue}&space;{{x}^{3}}y,\&space;{{x}^{2}}{{y}^{2}}}$  এবং  ${\color{Blue}&space;x{{y}^{3}}}$  ক্রমিক সমানুপাতী। সমাধানঃ $\frac{{{x}^{3}}y}{{x}^{2}{y}^{2}}$ $=\frac{x}{y}$ আবার, $\frac{x^{2}y^{2}}{x{{y}^{3}}}$ $=\frac{x}{y}$ উত্তরঃ বিবৃতিটি সত্য

 Q12. (C) শূন্যস্থান পূরণ করি : (i) তিনটি ক্রমিক সমানুপাতী ধনাত্মক সংখ্যার গুনফল 64 হলে, তাদের মধ্যসমানুপাতী _____। সমাধানঃ ধরি, তিনটি ধনাত্মক ক্রমিক সমানুপাতী সংখ্যা হল a, b, c  প্রদত্ত, $abc=64$ $\therefore&space;\frac{a}{b}=\frac{b}{c}$   বা, $b^{2}=ac$ বা, $b^{3}=abc$ বা, $b^{3}=64$ বা, $\left&space;(&space;b&space;\right&space;)^{3}=\left&space;(&space;4&space;\right&space;)^{3}$ $\therefore&space;b=4$ উত্তরঃ নির্ণেয় মধ্যসমানুপাতী 4

Koshe dekhi 5.3 class 10

 Q12. (C) (ii)  a : 2 = b : 5 = c : 8 হলে a এর 50% = b এর 20% = c এর ____ % সমাধানঃ প্রদত্ত, a : 2 = b : 5 = c : 8 বা,  $\frac{a}{2}=\frac{b}{5}=\frac{c}{8}$ ধরি, a এর 50% = b এর 20% = c এর x% বা, a এর $\frac{50}{100}$ = b এর $\frac{20}{100}$ = c এর $\frac{x}{100}$ বা,  $\frac{a}{2}=\frac{b}{5}=\frac{cx}{100}$ $\therefore&space;\frac{cx}{100}=\frac{c}{8}$ বা, $x=\frac{100}{8}$ $\therefore&space;x=12.5$ উত্তরঃ নির্ণেয় a এর 50% = b এর 20% = c এর 12.5%

Koshe dekhi 5.3 class 10

 Q12. (C) (iii) ${\color{Blue}&space;\left&space;(&space;x+2&space;\right&space;)}$  এবং ${\color{Blue}&space;\left&space;(&space;x-3&space;\right&space;)}$  -এর মধ্য সমানুপাতী x হলে, x -এর মান ______। সমাধানঃ $\left&space;(&space;x+2&space;\right&space;):x::x:\left&space;(&space;x-3&space;\right&space;)$ বা, $\frac{x+2}{x}=\frac{x}{x-3}$ বা, $\left&space;(&space;x+2&space;\right&space;)\left&space;(&space;x-3&space;\right&space;)=x^{2}$ বা, $x^{2}-3x+2x-6-x^{2}=0$ বা, $-x=6$ $\therefore&space;x=-6$ উত্তরঃ নির্ণেয় x -এর মান -6

Koshe dekhi 5.3 class 10

 Q13. সংক্ষিপ্ত উত্তরধর্মী প্রশ্ন (S.A.) (i) ${\color{Blue}&space;\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=\frac{2a-3b+4c}{p}}$  হলে,  p -এর মান নির্ণয় করি। সমাধানঃ প্রদত্ত, $\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=\frac{2a-3b+4c}{p}$ ধরি,  $\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=\frac{2a-3b+4c}{p}=k$ $\therefore&space;a=2k,b=3k,c=4k$  ও  $\left&space;(&space;2a-3b+4c&space;\right&space;)=pk$ a, b ও c এর মান বসিয়ে পাই, $2\times&space;2k-3\times&space;3k+4\times&space;4k=pk$ বা, $4k-9k+16k=pk$ বা, $11k=pk$ $\therefore&space;p=11$ উত্তরঃ নির্ণেয় p -এর মান 11

Koshe dekhi 5.3 class 10

 Q13 (ii) ${\color{Blue}&space;\frac{3x-5y}{3x+5y}=\frac{1}{2}}$  হলে,  ${\color{Blue}&space;\frac{3{{x}^{2}}-5{{y}^{2}}}{3{{x}^{2}}+5{{y}^{2}}}}$ -এর মান নির্ণয় করি। সমাধানঃ $\frac{3x-5y}{3x+5y}=\frac{1}{2}$ বা, $6x-10y=3x+5y$ বা, $6x-3x=5y+10y$ বা, $3x=15y$ বা, $\frac{x}{y}=\frac{15}{3}$ বা, $x=5y$ $\therefore&space;\frac{3{{x}^{2}}-5{{y}^{2}}}{3{{x}^{2}}+5{{y}^{2}}}$ -এর মান $=\frac{3\times&space;{{\left&space;(&space;5y&space;\right&space;)}^{2}}-5{{y}^{2}}}{3\times&space;{{\left&space;(&space;5y&space;\right&space;)}^{2}}+5{{y}^{2}}}$ $=\frac{75y^{2}-5{{y}^{2}}}{75y^{2}+5{{y}^{2}}}$ $=\frac{70y^{2}}{80y^{2}}$ $=\frac{7}{8}$ উত্তরঃ নির্ণেয়  ${\color{DarkGreen}&space;\frac{3{{x}^{2}}-5{{y}^{2}}}{3{{x}^{2}}+5{{y}^{2}}}}$ -এর মান ${\color{DarkGreen}&space;\frac{7}{8}}$

Koshe dekhi 5.3 class 10

 Q13 (iii)  a : b = 3 : 4  এবং  x : y = 5 : 7  হলে, ${\color{Blue}&space;\left(&space;3ax-by&space;\right):\left(&space;4by-7ax&space;\right)}$ কত নির্ণয় করি। সমাধানঃ প্রদত্ত, a : b = 3 : 4 বা, $\frac{a}{b}=\frac{3}{4}$ ধরি, a = 3k ও b=4k  এবং  x : y = 5 : 7 বা, $\frac{x}{y}=\frac{5}{7}$ ধরি, x=5l ও y=7l $\therefore&space;\left(&space;3ax-by&space;\right):\left(&space;4by-7ax&space;\right)$ $=\left(&space;3\times&space;3k\times&space;5l-4k\times&space;7l&space;\right):\left(&space;4\times&space;4k\times&space;7l&space;-7\times&space;3k\times&space;5l&space;\right)$ $=\left(&space;45kl-28kl\right):\left(&space;112kl-105kl&space;\right)$ $=17kl:7kl$ $=17:7$ উত্তরঃ নির্ণেয় ${\color{DarkGreen}&space;\left(&space;3ax-by&space;\right):\left(&space;4by-7ax&space;\right)=17:7}$

Koshe dekhi 5.3 class 10

 Q13 (iv) x, 12, y, 27  ক্রমিক সমানুপাতী হলে,  x ও y -এর ধনাত্মক মান নির্ণয় করি৷ সমাধানঃ প্রদত্ত, x, 12, y, 27  ক্রমিক সমানুপাতী $\therefore&space;\frac{x}{12}=\frac{12}{y}=\frac{y}{27}$ এখন, $\frac{12}{y}=\frac{y}{27}$ বা, $y^{2}=12\times&space;27$ বা, $y=\sqrt{12\times&space;27}$ $\therefore&space;y=18$ আবার, $\frac{x}{12}=\frac{12}{y}$ বা, $\frac{x}{12}=\frac{12}{18}$ বা, $18x=12\times&space;12$ বা, $x=\frac{12\times&space;12}{18}$ $\therefore&space;x=8$ উত্তরঃ নির্ণেয় x ও y -এর ধনাত্মক মান 8 ও 18

Koshe dekhi 5.3 class 10

 Q13 (v)  a : b = 3 : 2  এবং  b : c = 3 : 2  হলে,  a +b : b + c  কত নির্ণয় করি। সমাধানঃ প্রদত্ত, a : b = 3 : 2  এবং  b : c = 3 : 2 $\therefore&space;\frac{a}{b}=\frac{3}{2}$ বা, $\frac{a}{b}=\frac{3\times&space;3}{2\times&space;3}=\frac{9}{6}$ এবং   $\frac{b}{c}=\frac{3}{2}$ বা, $\frac{b}{c}=\frac{3\times&space;2}{2\times&space;2}=\frac{6}{4}$ এখন, a : b : c = 9 : 6 : 4 ধরি, $a=9k,b=6k,c=4k$ ∴ a +b : b + c $=\frac{a+b}{b+c}$ $=\frac{9k+6k}{6k+4k}$ $=\frac{15k}{10k}$ $=\frac{3}{2}$ উত্তরঃ নির্ণেয় a +b : b + c = 3 : 2

Koshe dekhi 5.3 class 10

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