Sat. Jun 22nd, 2024

Koshe Dekhi 2 Class 9

Koshe Dekhi 2 Class 9

1. মান নির্ণয় করি :

(i) \fn_cm {\color{Blue} \left ( \sqrt[5]{8} \right )^{\frac{5}{2}}\times \left ( 16 \right )^{\frac{-3}{2}}}  

সমাধানঃ 

\fn_cm \left ( \sqrt[5]{8} \right )^{\frac{5}{2}}\times \left ( 16 \right )^{\frac{-3}{2}}

=\left ( \sqrt[5]{2^{3}} \right )^{\frac{5}{2}}\times \left ( 2^{4} \right )^{-\frac{3}{2}}

=2^{\frac{3}{5}\times \frac{5}{2}}\times \frac{1}{2^{4\times \frac{3}{2}}}

=2^{\frac{3}{2}}\times \frac{1}{2^{6}}

=2^{\frac{3}{2}}\times 2^{-6}

=2^{\frac{3}{2}-6}

=2^{-\frac{9}{2}}

=\frac{1}{2^{\frac{9}{2}}}

=2^{-\frac{9}{2}}

 

1. মান নির্ণয় করি :

(ii) \fn_cm {\color{Blue} \left \{ \left ( 125 \right )^{-2}\times \left ( 16 \right )^{\frac{-3}{2}} \right \}^{\frac{-1}{6}}} 

সমাধানঃ 

\fn_cm \left \{ \left ( 125 \right )^{-2}\times \left ( 16 \right )^{\frac{-3}{2}} \right \}^{\frac{-1}{6}}

\fn_cm =\left \{ \left ( 5^{3} \right )^{-2}\times \left ( 2^{4} \right )^{\frac{-3}{2}} \right \}^{\frac{-1}{6}}

=\left \{ 5^{-6}\times 2^{-6} \right \}^{\frac{-1}{6}}

=\left (5^{-6} \right )^{-\frac{1}{6}}\times \left (2^{-6} \right )^{-\frac{1}{6}}

= 5 × 2

= 10

 

1. মান নির্ণয় করি :

(iii) \fn_cm {\color{Blue} 4^{\frac{1}{3}}\times \left [ 2^{\frac{1}{3}}\times 3^{\frac{1}{2}} \right ]\div 9^{\frac{1}{4}}} 

সমাধানঃ 

\fn_cm 4^{\frac{1}{3}}\times \left [ 2^{\frac{1}{3}}\times 3^{\frac{1}{2}} \right ]\div 9^{\frac{1}{4}}

\fn_cm =\left (2^{2} \right )^{\frac{1}{3}}\times \left [ 2^{\frac{1}{3}}\times 3^{\frac{1}{2}} \right ]\div \left ( 3^{2} \right )^{\frac{1}{4}}

\fn_cm =2^{\frac{2}{3}}\times \left [ 2^{\frac{1}{3}}\times 3^{\frac{1}{2}} \right ]\div 3^{\frac{1}{2}}

\fn_cm =2^{\frac{2}{3}}\times \left [ 2^{\frac{1}{3}}\times 3^{\frac{1}{2}} \right ]\times \frac{1}{3^{\frac{1}{2}}}

\fn_cm =2^{\frac{2}{3}}\times 2^{\frac{1}{3}}\times 3^{\frac{1}{2}} \times 3^{-\frac{1}{2}}

\fn_cm =2^{\frac{2}{3}+{\frac{1}{3}}}\times 3^{\frac{1}{2}-\frac{1}{2}}

=\left (2 \right )^{\frac{2+1}{3}}\times \left (3 \right )^{\frac{1-1}{2}}

=2^{1}\times 3^{0}

= 2 × 1

= 2

 

2. সরল করি :

(i) \fn_cm {\color{Blue} \left ( 8a^{3} \div 27x^{-3} \right )^{\frac{2}{3}} \times\left ( 64a^{3} \div 27x^{-3} \right )^{\frac{-2}{3}}} 

সমাধানঃ 

\fn_cm \left ( 8a^{3} \div 27x^{-3} \right )^{\frac{2}{3}} \times\left ( 64a^{3} \div 27x^{-3} \right )^{\frac{-2}{3}}

\fn_cm =\left [\left ( 2a \right )^{3\times \frac{2}{3}} \div \left ( 3x \right )^{-3\times \frac{2}{3}} \right ]\times \left [\left ( 4a \right )^{3\times \frac{-2}{3}} \div \left ( 3x \right )^{-3\times \frac{-2}{3}} \right ]

=\left [ \left (2a \right )^{2} \div \left ( 3x \right )^{-2} \right ]\times \left [ \left ( 4a \right )^{-2} \div \left ( 3x \right )^{2} \right ]

=4a^{2}\times \frac{1}{\frac{1}{9x^{2}}}\times \frac{1}{16a^{2}}\times \frac{1}{9x^{2}}

=\frac{1}{4}\times 1\times {\frac{9x^{2}}{1}}\times \frac{1}{9x^{2}}

=\frac{1}{4}

 

2. সরল করি :

(ii) \fn_cm {\color{Blue} \left \{ \left ( x^{-5} \right )^{\frac{2}{3}} \right \}^{\frac{-3}{10}}} 

সমাধানঃ 

\fn_cm \left \{ \left ( x^{-5} \right )^{\frac{2}{3}} \right \}^{\frac{-3}{10}}

=\left \{ \left ( \frac{1}{x} \right )^{5\times \frac{2}{3}} \right \}^{\frac{-3}{10}}

=\left \{ \left ( \frac{1}{x} \right )^{\frac{10}{3}} \right \}^{\frac{-3}{10}}

=\left \{ \left ( \frac{1}{x} \right )^{\frac{10}{3}\times \frac{-3}{10}} \right \}

=\left ( \frac{1}{x} \right )^{-1}

=\frac{1}{\frac{1}{x}}

=1\times \frac{x}{1}

=x

 

2. সরল করি :

(iii) \fn_cm {\color{Blue} \left [ \left \{ \left ( 2^{-1} \right )^{-1} \right \}^{-1} \right ]^{-1}} 

সমাধানঃ 

\fn_cm \left [ \left \{ \left ( 2^{-1} \right )^{-1} \right \}^{-1} \right ]^{-1}

\fn_cm =\left [ \left \{ \left ( \frac{1}{2} \right )^{-1} \right \}^{-1} \right ]^{-1}

\fn_cm =\left [ \left \{ 2 \right \}^{-1} \right ]^{-1}

\fn_cm =\left [ \frac{1}{2}\right ]^{-1}

= 2

 

2. সরল করি :

(iv) \fn_cm {\color{Blue} \sqrt[3]{a^{-2}}.b\times \sqrt[3]{b^{-2}}.c\times \sqrt[3]{c^{-2}}.a} 

সমাধানঃ 

\fn_cm \sqrt[3]{a^{-2}}.b\times \sqrt[3]{b^{-2}}.c\times \sqrt[3]{c^{-2}}.a

=a^{\frac{-2}{3}}b\times b^{\frac{-2}{3}}c\times c^{\frac{-2}{3}}a

=a^{\frac{-2}{3}+1}\times b^{\frac{-2}{3}+1}\times c^{\frac{-2}{3}+1}

=a^{\frac{1}{3}}\times b^{\frac{1}{3}}\times c^{\frac{1}{3}}

=\left (abc \right )^{\frac{1}{3}}

=\sqrt[3]{abc}

 

2. সরল করি :

(v) \fn_cm {\color{Blue} \left ( \frac{4^{m+\frac{1}{4}}\times \sqrt{2.2^{m}}}{2.\sqrt{2^{-m}}} \right )^{\frac{1}{m}}} 

সমাধানঃ 

\fn_cm \left ( \frac{4^{m+\frac{1}{4}}\times \sqrt{2.2^{m}}}{2.\sqrt{2^{-m}}} \right )^{\frac{1}{m}}

\fn_cm =\left ( \frac{\left (2^{2} \right )^{m+\frac{1}{4}}\times \sqrt{2.2^{m}}}{2\times 2^{\frac{-m}{2}}} \right )^{\frac{1}{m}}

=\left (\frac{2^{2m}\times 2^{2\times \frac{1}{4}}\times 2^{\frac{1}{2}}\times 2^{\frac{m}{2}}}{2.2^{\frac{-m}{2}}} \right )^{\frac{1}{m}}

=\left (\frac{2^{2m}\times 2^{\frac{1}{2}}\times 2^{\frac{1}{2}}\times 2^{\frac{m}{2}}}{2.2^{\frac{-m}{2}}} \right )^{\frac{1}{m}}

=\left (\frac{2^{2m}\times 2\times 2^{\frac{m}{2}}}{2.2^{\frac{-m}{2}}} \right )^{\frac{1}{m}}

=\left (2^{2m}\times 2^{\frac{m}{2}}\times 2^{\frac{m}{2}} \right )^{\frac{1}{m}}

=\left (2^{2m+\frac{m}{2}+\frac{m}{2}}\right )^{\frac{1}{m}}

=\left (2^{\frac{6m}{2}}\right )^{\frac{1}{m}}

=\left (2^{3m} \right )^{\frac{1}{m}}

\fn_cm =2^{3}

= 8

 

2. সরল করি :

(vi) \fn_cm {\color{Blue} 9^{-3}\times \frac{16^{\frac{1}{4}}}{6^{-2}}\times \left ( \frac{1}{27} \right )^{\frac{-4}{3}}}

সমাধানঃ 

\fn_cm 9^{-3}\times \frac{16^{\frac{1}{4}}}{6^{-2}}\times \left ( \frac{1}{27} \right )^{\frac{-4}{3}}

\fn_cm =\left ( 3^{2} \right )^{-3}\times \frac{\left ( 2^{4} \right )^{\frac{1}{4}}}{\left ( 2\times 3 \right )^{-2}}\times \left ( \frac{1}{3^{3}} \right )^{\frac{-4}{3}}

\fn_cm =\left ( 3 \right )^{-6}\times \frac{\left ( 2 \right )^{4\times \frac{1}{4}}}{2^{-2}\times 3^{-2}}\times \left ( \frac{1}{3^{3\times \frac{-4}{3}}} \right )

=3^{-6}\times 2\times 2^{2}\times 3^{2}\times 3^{4}

=3^{-6+6}\times 2^{1+2}

=3^{0}\times 2^{3}

= 1 × 8

= 8

 

2. সরল করি :

(vii) \fn_cm {\color{Blue} \left ( \frac{x^{a}}{x^{b}} \right )^{a^{2}+ab+b^{2}}\times \left ( \frac{x^{b}}{x^{c}} \right )^{b^{2}+bc+c^{2}}\times \left ( \frac{x^{c}}{x^{a}} \right )^{c^{2}+ca+a^{2}}}

সমাধানঃ 

\fn_cm \left ( \frac{x^{a}}{x^{b}} \right )^{a^{2}+ab+b^{2}}\times \left ( \frac{x^{b}}{x^{c}} \right )^{b^{2}+bc+c^{2}}\times \left ( \frac{x^{c}}{x^{a}} \right )^{c^{2}+ca+a^{2}}

=\left ( \frac{x^{a^{3}+a^{2}b+ab^{2}}}{x^{a^{2}b+ab^{2}+b^{3}}} \right )\times \left ( \frac{x^{b^{3}+b^{2}c+bc^{2}}}{x^{b^{2}c+bc^{2}+c^{3}}} \right )\times \left ( \frac{x^{c^{3}+c^{2}a+ca^{2}}}{x^{ac^{2}+ca^{2}+a^{3}}} \right )

=\left ( x^{a^{3}+a^{2}b+ab^{2}-{a^{2}b-ab^{2}-b^{3}+b^{3}+b^{2}c+bc^{2}-b^{2}c-bc^{2}-c^{3}+c^{3}+c^{2}a+ca^{2}-ac^{2}-ca^{2}-a^{3}}} \right )

=x^{0}

= 1

 

3. মানের ঊর্ধ্বক্রমানুসারে সাজাই :

(i) \fn_cm {\color{Blue} 5^{\frac{1}{2}},10^{\frac{1}{4}},6^{\frac{1}{3}}} 

সমাধানঃ 

\fn_cm 5^{\frac{1}{2}},10^{\frac{1}{4}},6^{\frac{1}{3}}

উপরের সংখ্যাগুলির সূচকগুলি হলো  \fn_cm \frac{1}{2},\frac{1}{4}\fn_cm \frac{1}{3}

2, 4 ও 3 -এর ল.সা.গু. = 12

 

\fn_cm \frac{1}{2}=\frac{1\times 6}{2\times 6}=\frac{6}{12}

\fn_cm \frac{1}{4}=\frac{1\times 3}{4\times 3}=\frac{3}{12}

\fn_cm \frac{1}{3}=\frac{1\times 4}{3\times 4}=\frac{4}{12}

 

এখন, \fn_cm 5^{\frac{1}{2}}=5^{\frac{6}{12}}=\sqrt[12]{5^{6}}=\sqrt[12]{15625}

\fn_cm 10^{\frac{1}{4}}=10^{\frac{3}{12}}=\sqrt[12]{10^{3}}=\sqrt[12]{1000}

\fn_cm 6^{\frac{1}{3}}=6^{\frac{4}{12}}=\sqrt[12]{6^{4}}=\sqrt[12]{1296}

 

আমরা দেখতে পাচ্ছি যে, \fn_cm \sqrt[12]{1000}<\sqrt[12]{1296}<\sqrt[12]{15625}

অর্থাৎ, মানের উর্ধক্রমে সাজিয়ে পাই \fn_cm 10^{\frac{1}{4}}<6^{\frac{1}{3}}<5^{\frac{1}{2}} (উত্তর)

 

3. মানের ঊর্ধ্বক্রমানুসারে সাজাই :

(ii) \fn_cm {\color{Blue} 3^{\frac{1}{3}},2^{\frac{1}{2}},8^{\frac{1}{4}}} 

সমাধানঃ 

\fn_cm 3^{\frac{1}{3}},2^{\frac{1}{2}},8^{\frac{1}{4}}

উপরের সংখ্যাগুলির সূচকগুলি হলো   \fn_cm \frac{1}{3} , \fn_cm \frac{1}{2}\frac{1}{4}

3, 2 ও 4 -এর ল.সা.গু. = 12

 

\fn_cm \frac{1}{3}=\frac{1\times 4}{3\times 4}=\frac{4}{12}

\fn_cm \frac{1}{2}=\frac{1\times 6}{2\times 6}=\frac{6}{12}

\fn_cm \frac{1}{4}=\frac{1\times 3}{4\times 3}=\frac{3}{12}

 

এখন, \fn_cm 3^{\frac{1}{3}}=3^{\frac{4}{12}}=\sqrt[12]{3^{4}}=\sqrt[12]{81}

\fn_cm 2^{\frac{1}{2}}=2^{\frac{6}{12}}=\sqrt[12]{2^{6}}=\sqrt[12]{64}

\fn_cm 8^{\frac{1}{4}}=8^{\frac{3}{12}}=\sqrt[12]{8^{3}}=\sqrt[12]{512}

 

আমরা দেখতে পাচ্ছি যে, \fn_cm \sqrt[12]{64}<\sqrt[12]{81}<\sqrt[12]{512}

অর্থাৎ, মানের উর্ধক্রমে সাজিয়ে পাই \fn_cm 2^{\frac{1}{2}}<3^{\frac{1}{3}}<8^{\frac{1}{4}} (উত্তর)

 

3. মানের ঊর্ধ্বক্রমানুসারে সাজাই :

(iii) \fn_cm {\color{Blue} 2^{60},3^{48},4^{36},5^{24}} 

সমাধানঃ 

\fn_cm 2^{60},3^{48},4^{36},5^{24}

উপরের সংখ্যাগুলির সূচকগুলি হলো   60 , 48 , 36 ও 24

60, 48, 36 ও 24 -এর গ.সা.গু. = 12

 

\fn_cm \frac{1}{3}=\frac{1\times 4}{3\times 4}=\frac{4}{12}

\fn_cm \frac{1}{2}=\frac{1\times 6}{2\times 6}=\frac{6}{12}

\fn_cm \frac{1}{4}=\frac{1\times 3}{4\times 3}=\frac{3}{12}

 

এখন, \fn_cm 2^{60}=2^{\frac{60}{12}}=2^{5}=32

\fn_cm 3^{48}=3^{\frac{48}{12}}=3^{4}=81

\fn_cm 4^{36}=4^{\frac{36}{12}}=4^{3}=48

\fn_cm 5^{24}=5^{\frac{24}{12}}=5^{2}=25

আমরা দেখতে পাচ্ছি যে, 25<32<48<81

অর্থাৎ, মানের উর্ধক্রমে সাজিয়ে পাই 5^{24}<2^{60}<4^{36}<3^{48} (উত্তর)

Koshe dekhi 2 Class 9 

 

4. প্রমান করি :

(i) \fn_cm {\color{Blue} \left ( \frac{a^{q}}{a^{r}} \right )^{p}\times \left ( \frac{a^{r}}{a^{p}} \right )^{q}\times \left ( \frac{a^{p}}{a^{q}} \right )^{r}=1} 

সমাধানঃ

বামপক্ষ : 

\fn_cm \left ( \frac{a^{q}}{a^{r}} \right )^{p}\times \left ( \frac{a^{r}}{a^{p}} \right )^{q}\times \left ( \frac{a^{p}}{a^{q}} \right )^{r}

=\frac{a^{pq}}{a^{pr}}\times \frac{a^{qr}}{a^{pq}}\times \frac{a^{pr}}{a^{qr}}

= 1 

= ডানপক্ষ (প্রমাণিত)

 

4. প্রমান করি :

(ii) \fn_cm {\color{Blue} \left ( \frac{x^{m}}{x^{n}} \right )^{m+n}\times \left ( \frac{x^{n}}{x^{l}} \right )^{n+l}\times \left ( \frac{x^{l}}{x^{m}} \right )^{l+m}=1} 

সমাধানঃ 

বামপক্ষ : 

\fn_cm \left ( \frac{x^{m}}{x^{n}} \right )^{m+n}\times \left ( \frac{x^{n}}{x^{l}} \right )^{n+l}\times \left ( \frac{x^{l}}{x^{m}} \right )^{l+m}

=\frac{x^{m^{2}+mn}}{x^{mn+n^{2}}}\times \frac{x^{n^{2}+nl}}{x^{nl+l^{2}}}\times \frac{x^{l^{2}+lm}}{x^{lm+m^{2}}}

=x^{m^{2}+mn-mn-n^{2}+n^{2}+nl-nl-l^{2}+l^{2}+lm-lm-m^{2}}

=x^{0}

= 1 

= ডানপক্ষ (প্রমাণিত)

 

4. প্রমান করি :

(iii) \fn_cm {\color{Blue} \left ( \frac{x^{m}}{x^{n}} \right )^{m+n-l}\times \left ( \frac{x^{n}}{x^{l}} \right )^{n+l-m}\times \left ( \frac{x^{l}}{x^{m}} \right )^{l+m-n}=1} 

সমাধানঃ 

বামপক্ষ : 

\fn_cm \left ( \frac{x^{m}}{x^{n}} \right )^{m+n-l}\times \left ( \frac{x^{n}}{x^{l}} \right )^{n+l-m}\times \left ( \frac{x^{l}}{x^{m}} \right )^{l+m-n}

=\frac{x^{m^{2}+mn-ml}}{x^{mn+n^{2}-nl}}\times \frac{x^{n^{2}+nl-mn}}{x^{nl+l^{2}-lm}}\times \frac{x^{l^{2}+lm-nl}}{x^{lm+m^{2}-mn}}

=x^{m^{2}+mn-ml-mn-n^{2}+nl+n^{2}+nl-mn-nl-l^{2}+lm+l^{2}+lm-nl-lm-m^{2}+mn}

=x^{0}

= 1 

= ডানপক্ষ (প্রমাণিত)

 

4. প্রমান করি :

(iv) \fn_cm {\color{Blue} \left ( a^{\frac{1}{x-y}} \right )^{\frac{1}{x-z}}\times \left ( a^{\frac{1}{y-z}} \right )^{\frac{1}{y-x}}\times \left ( a^{\frac{1}{z-x}} \right )^{\frac{1}{z-y}}=1} 

সমাধানঃ 

বামপক্ষ : 

\fn_cm \left ( a^{\frac{1}{x-y}} \right )^{\frac{1}{x-z}}\times \left ( a^{\frac{1}{y-z}} \right )^{\frac{1}{y-x}}\times \left ( a^{\frac{1}{z-x}} \right )^{\frac{1}{z-y}}

\fn_cm =\left ( a^{\frac{1}{x-y}\times \frac{1}{x-z}} \right )\times \left ( a^{\frac{1}{y-z}\times \frac{1}{y-x}} \right )\times \left ( a^{\frac{1}{z-x}\times \frac{1}{z-y}} \right )

\fn_cm =\left [ a^{\frac{1}{x-y}\times \frac{1}{x-z}} \right ]\times \left [ a^{\frac{1}{y-z}\times \frac{1}{-\left ( x-y \right )}} \right ]\times \left [ a^{\frac{1}{-\left ( x-z \right )}\times \frac{1}{-\left ( y-z \right )}} \right ]

\fn_cm =\left ( a^{\frac{1}{x-y}\times \frac{1}{x-z}-\frac{1}{y-z}\times \frac{1}{x-y}+\frac{1}{x-z}\times \frac{1}{y-z}} \right )

=a^{\frac{y-z-x+z+x-y}{\left ( x-y \right )\left ( x-z \right )\left ( y-z \right )}}

=a^{\frac{0}{\left ( x-y \right )\left ( x-z \right )\left ( y-z \right )}}

=a^{0}

= 1 

= ডানপক্ষ (প্রমাণিত)

 

5. \fn_cm {\color{Blue} x+z=2y} এবং \fn_cm {\color{Blue} b^{2}=ac} হলে, দেখাই যে, \fn_cm {\color{Blue} a^{y-z}b^{z-x}c^{x-y}=1}

সমাধানঃ 

প্রদত্ত,

x + z = 2y 

বা, x – y = y – z 

এবং,

\fn_cm b^{2}=ac

বা, b=\sqrt{ac}

বামপক্ষ : 

\fn_cm a^{y-z}b^{z-x}c^{x-y}

\fn_cm =a^{x-y}\left ( \sqrt{ac} \right )^{z-x}c^{x-y}

\fn_cm =\left (ac \right )^{x-y}\left ( \sqrt{ac} \right )^{z-x}

=a^{x-y}\times c^{x-y}\times a^{\frac{z-x}{2}}\times c^{\frac{z-x}{2}}

=a^{x-y+\frac{z-x}{2}}\times c^{x-y+\frac{z-x}{2}}

=a^{\frac{2x-2y+z-x}{2}}\times c^{\frac{2x-2y+z-x}{2}}

=a^{\frac{x+z-2y}{2}}\times c^{\frac{x+z-2y}{2}}

=a^{\frac{2y-2y}{2}}\times c^{\frac{2y-2y}{2}}

=a^{\frac{0}{2}}\times c^{\frac{0}{2}}

=a^{0}\times c^{0}

= 1 × 1

= 1 

= ডানপক্ষ (প্রমাণিত)

 

6. \fn_cm {\color{Blue} a=xy^{p-1},b=xy^{q-1}} এবং \fn_cm {\color{Blue} c=xy^{r-1}} হলে, দেখাই যে, \fn_cm {\color{Blue} a^{q-r}b^{r-p}c^{p-q}=1}

সমাধানঃ 

বামপক্ষ : 

\fn_cm a^{q-r}b^{r-p}c^{p-q}

\fn_cm =\left ( xy^{p-1} \right )^{q-r}\left ( xy^{q-1} \right )^{r-p}\left ( xy^{r-1} \right )^{p-q}

\fn_cm =\left [ x^{q-r}\times y^{\left (p-1 \right )\left (q-r \right )} \right ]\left [ x^{r-p}\times y^{\left (q-1 \right )\left (r-p \right )} \right ]\left [ x^{p-q}\times y^{\left (p-q \right )\left (r-1 \right )} \right ]

\fn_cm =\left [ x^{q-r+r-p+p-q}\times y^{\left (p-1 \right )\left (q-r \right )+\left (q-1 \right )\left (r-p \right )+\left (p-q \right )\left (r-1 \right )} \right ]

=x^{0}\times y^{pq-pr-q+r+qr-pq-r+p+pr-p-qr+q}

=x^{0}\times y^{0}

= 1 × 1

= 1 

= ডানপক্ষ (প্রমাণিত)

 

7. \fn_cm {\color{Blue} x^{\frac{1}{a}}=y^{\frac{1}{b}}=z^{\frac{1}{c}}} এবং \fn_cm {\color{Blue} xyz=1} হলে, দেখাই যে, \fn_cm {\color{Blue} a+b+c=0}.

সমাধানঃ 

\fn_cm x^{\frac{1}{a}}=y^{\frac{1}{b}}=z^{\frac{1}{c}}

বা, \fn_cm x^{\frac{1}{a}}=y^{\frac{1}{b}}=z^{\frac{1}{c}}=k ( ধরি )

\therefore x=k^{a},y=k^{b},z=k^{c}

এখন,

xyz=1

বা, k^{a}\times k^{b}\times k^{c}=1

বা, k^{a+b+c}=1

বা, k^{a+b+c}=k^{0}

\therefore a+b+c=0 (প্রমানিত)

Koshe dekhi 2 Class 9 

 

8. \fn_cm {\color{Blue} x^{a}=b^{y}=z^{c}} এবং \fn_cm {\color{Blue} abc=1} হলে, দেখাই যে, \fn_cm {\color{Blue} xy+yz+zx=0.}

সমাধানঃ 

\fn_cm x^{a}=b^{y}=z^{c}

বা, \fn_cm x^{a}=b^{y}=z^{c}=k ( ধরি )

\therefore x=k^{\frac{1}{a}},b=k^{\frac{1}{y}},z=k^{\frac{1}{c}}

আবার,

abc=1

এখন,

\fn_cm xy+yz+zx

=k^{\frac{1}{a}}y+yk^{\frac{1}{c}}+k^{\frac{1}{c}}k^{\frac{1}{a}}

বা, k^{a}\times k^{b}\times k^{c}=1

বা, k^{a+b+c}=1

বা, k^{a+b+c}=k^{0}

\therefore a+b+c=0 (প্রমানিত)

 

9. সমাধান করি 

(i) \fn_cm {\color{Blue} 49^{x}=7^{3}} 

সমাধানঃ 

\fn_cm 49^{x}=7^{3}

বা, \left (7^{2} \right )^{x}=7^{3}

বা, 7^{2x}=7^{3}

বা, 2x=3

\therefore x=\frac{3}{2}

নির্ণেয় x এর মান {\color{DarkGreen} \frac{3}{2}}

 

9. সমাধান করি 

(ii) \fn_cm {\color{Blue} 2^{x+2}+2^{x-1}=9} 

সমাধানঃ 

\fn_cm 2^{x+2}+2^{x-1}=9

বা, \fn_cm 2^{x}\times 2^{2}+2^{x}\times 2^{-1}=9

বা, \fn_cm 2^{x}\left ( 4+\frac{1}{2} \right )=9

বা, \fn_cm 2^{x}\times \frac{9}{2}=9

বা, \fn_cm 2^{x}=9\times \frac{2}{9}

বা, \fn_cm 2^{x}=2^{1}

\therefore x=1

নির্ণেয় x এর মান 1 

 

9. সমাধান করি 

(iii) \fn_cm {\color{Blue} 2^{x+1}+2^{x+2}=48} 

সমাধানঃ 

\fn_cm 2^{x+1}+2^{x+2}=48

বা, \fn_cm 2^{x}\times 2^{1}+2^{x}\times 2^{2}=48

বা, \fn_cm 2^{x}\left ( 2+4 \right )=48

বা, \fn_cm 2^{x}\times 6=48

বা, \fn_cm 2^{x}=48\times \frac{1}{6}

বা, \fn_cm 2^{x}=8

বা, \fn_cm 2^{x}=2^{3}

\therefore x=3

নির্ণেয় x এর মান 3

 

9. সমাধান করি 

(iv) \fn_cm {\color{Blue} 2^{4x}.4^{3x-1}=\frac{4^{2x}}{2^{3x}}} 

সমাধানঃ 

\fn_cm 2^{4x}.4^{3x-1}=\frac{4^{2x}}{2^{3x}}

বা, \fn_cm 2^{4x}.\left ( 2^{2} \right )^{3x-1}=\frac{\left ( 2^{2} \right )^{2x}}{2^{3x}}

বা, \fn_cm 2^{4x}\times 2^{6x-2}=2^{4x-3x}

বা, 2^{4x+6x-2}=2^{x}

বা, 2^{10x-2}=2^{x}

বা, 2^{10x}\times 2^{-2}=2^{x}

বা, \frac{2^{10x}}{4}=2^{x}

বা, \frac{2^{10x}}{2^{x}}=2^{2}

বা, 2^{10x-x}=2^{2}

বা, 9x=2

\therefore x=\frac{2}{9}

নির্ণেয় x এর মান \fn_cm {\color{DarkGreen} \frac{2}{9}}

 

9. সমাধান করি 

(v) \fn_cm {\color{Blue} 9\times 81^{x}=27^{2-x}} 

সমাধানঃ 

\fn_cm 9\times 81^{x}=27^{2-x}

বা, \fn_cm \left ( 3^{2} \right )\times \left ( 3^{4} \right )^{x}=\left ( 3^{3} \right )^{2-x}

বা, \fn_cm 3^{2}\times 3^{4x}=3^{6-3x}

বা, \fn_cm 3^{2+4x}=3^{6-3x}

বা, \fn_cm 2+4x=6-3x

বা, 4x+3x=6-2

বা, 7x=4

\therefore x=\frac{4}{7}

নির্ণেয় x এর মান {\color{DarkGreen} \frac{4}{7}}

 

9. সমাধান করি 

(vi) \fn_cm {\color{Blue} 2^{5x+4}+2^{9}=2^{10}}

সমাধানঃ 

\fn_cm 2^{5x+4}+2^{9}=2^{10}

বা, \fn_cm 2^{5x}\times 2^{4}+2^{9}=2^{10}

বা, \fn_cm 2^{5x}\times 2^{4}=2^{10}-2^{9}

বা, \fn_cm 2^{5x}\times 2^{4}=2^{9}\left ( 2-1 \right )

বা, \fn_cm 2^{5x}=\frac{2^{9}}{2^{4}}

বা, \fn_cm 2^{5x}=2^{5}

বা, \fn_cm 5x=5

{\color{DarkGreen} \therefore x=\frac{5}{5}=1}

 

9. সমাধান করি 

(vii) \fn_cm {\color{Blue} 6^{2x+4}=3^{3x}.2^{x+8}}

সমাধানঃ 

\fn_cm 6^{2x+4}=3^{3x}.2^{x+8}

বা, \fn_cm \left ( 3\times 2 \right )^{2x+4}=3^{3x}.2^{x+8}

বা, \fn_cm 3^{2x+4}\times 2^{2x+4}=3^{3x}.2^{x+8}

বা, \fn_cm \frac{3^{2x+4}}{3^{3x}}=\frac{2^{x+8}}{2^{2x+4}}

বা, \fn_cm 3^{2x+4-3x}=2^{x+8-2x-4}

বা, \fn_cm 3^{4-x}=2^{4-x}

বা, \fn_cm \frac{3^{4}}{3^{x}}=\frac{2^{4}}{2^{x}}

বা, \fn_cm \frac{3^{4}}{2^{4}}=\frac{3^{x}}{2^{x}}

বা, \fn_cm \left (\frac{3}{2} \right )^{4}=\left (\frac{3}{2} \right )^{x}

{\color{DarkGreen} \therefore x=4}

Koshe dekhi 2 Class 9 

 

10. বহু বিকল্পীয় প্রশ্ন (M.C.Q.) :

(i) \fn_cm {\color{Blue} \left ( 0.243 \right )^{0.2}\times \left ( 10 \right )^{0.6}} এর মান –

(a) 0.3

(b) 3

(c) 0.9

(d) 9

সমাধানঃ 

\fn_cm \left ( 0.243 \right )^{0.2}\times \left ( 10 \right )^{0.6}

\fn_cm =\left ( \frac{243}{1000}\right )^{\frac{2}{10}}\times \left ( 10 \right )^{\frac{6}{10}}

\fn_cm =\left [ \frac{3^{5}}{\left (10 \right )^{3}}\right ]^{\frac{1}{5}}\times \left ( 10 \right )^{\frac{3}{5}}

\fn_cm =\left [ \frac{3^{5\times \frac{1}{5}}}{\left (10 \right )^{3\times \frac{1}{5}}}\right ]\times \left ( 10 \right )^{\frac{3}{5}}

\fn_cm =\left [ \frac{3}{\left (10 \right )^{\frac{3}{5}}}\right ]\times \left ( 10 \right )^{\frac{3}{5}}

=3

উত্তরঃ (b) 3

 

10. বহু বিকল্পীয় প্রশ্ন (M.C.Q.) :

(ii) \fn_cm {\color{Blue} 2^{\frac{1}{2}}\times 2^{-\frac{1}{2}}\times \left ( 16 \right )^{\frac{1}{2}}} এর মান –

(a) 1

(b) 2

(c) 4

(d) \fn_cm {\color{Blue} \frac{1}{2}}

সমাধানঃ 

\fn_cm 2^{\frac{1}{2}}\times 2^{-\frac{1}{2}}\times \left ( 16 \right )^{\frac{1}{2}}

\fn_cm =2^{\frac{1}{2}}\times 2^{-\frac{1}{2}}\times \left ( 2^{4} \right )^{\frac{1}{2}}

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

\fn_cm =2^{\frac{1}{2}-\frac{1}{2}+2}

=2^{2}

 = 4

উত্তরঃ (c) 4

 

10. বহু বিকল্পীয় প্রশ্ন (M.C.Q.) :

(iii) \fn_cm {\color{Blue} 4^{x}=8^{3}} হলে, x -এর মান –

(a) \fn_cm {\color{Blue} \frac{3}{2}}

(b) \fn_cm {\color{Blue} \frac{9}{2}}

(c) 3

(d) 9

 সমাধানঃ 

\fn_cm 4^{x}=8^{3}

বা, \left ( 2^{2} \right )^{x}=\left ( 2^{3} \right )^{3}

বা, 2^{2x}=2^{9}

বা, 2x=9

\therefore x=\frac{9}{2}

উত্তরঃ (b) \fn_cm {\color{DarkGreen} \frac{9}{2}}

 

10. বহু বিকল্পীয় প্রশ্ন (M.C.Q.) :

(iv) \fn_cm {\color{Blue} 20^{-x}=\frac{1}{7}} হলে, \fn_cm {\color{Blue} \left ( 20 \right )^{2x}} -এর মান –

(a) \fn_cm {\color{Blue} \frac{1}{49}}

(b) 7

(c) 49

(d) 1

 সমাধানঃ 

\fn_cm 20^{-x}=\frac{1}{7}

বা, \frac{1}{20^{x}}=\frac{1}{7}

বা, 20^{x}=7

বা, \left (20^{x} \right )^{2}=7^{2}

\therefore 20^{2x}=49

উত্তরঃ (c) 49

 

10. বহু বিকল্পীয় প্রশ্ন (M.C.Q.) :

(v) \fn_cm {\color{Blue} 4\times 5^{x}=500} হলে, \fn_cm {\color{Blue} x^{x}} -এর মান –

(a) 8

(b) 1

(c) 64

(d) 27

 সমাধানঃ 

\fn_cm 4\times 5^{x}=500

বা, \fn_cm 4\times 5^{x}=4\times 5^{3}

বা, \fn_cm 5^{x}=5^{3}

\therefore x=3

\fn_cm \therefore x^{x} -এর মান =3^{3}=27

উত্তরঃ (d) 27

Koshe dekhi 2 Class 9 

 

11. সংক্ষিপ্ত উত্তরভিত্তিক প্রশ্নঃ 

(i) \fn_cm {\color{Blue} \left ( 27 \right )^{x}=\left ( 81 \right )^{y}}  হলে, \fn_cm {\color{Blue} x:y} কত হয় লিখি।

সমাধানঃ 

\fn_cm \left ( 27 \right )^{x}=\left ( 81 \right )^{y}

বা, \fn_cm \left ( 3^{3} \right )^{x}=\left ( 3^{4} \right )^{y}

বা, \fn_cm \left ( 3 \right )^{3x}=\left ( 3 \right )^{4y}

বা, 3x=4y

বা, \frac{x}{y}=\frac{4}{3}

{\color{DarkGreen} \therefore x:y=4:3}

 

11. সংক্ষিপ্ত উত্তরভিত্তিক প্রশ্নঃ 

(ii) \fn_cm {\color{Blue} \left ( 5^{5}+0.01 \right )^{2}-\left ( 5^{5}-0.01 \right )^{2}=5^{x}} হলে, \fn_cm {\color{Blue} x} -এর মান কত হিসাব করে লিখি।

সমাধানঃ 

\fn_cm \left ( 5^{5}+0.01 \right )^{2}-\left ( 5^{5}-0.01 \right )^{2}=5^{x}

ধরি, {\color{Purple} 5^{5}=a,0.01=b}

বা, \fn_cm 4\times 5^{5}\times 0.01=5^{x}{\color{Purple} \left [ \because \left ( a+b \right )^{2} -\left ( a-b \right )^{2}=4ab\right ]}

বা, \fn_cm 4\times 5^{5}\times \frac{1}{100}=5^{x}

বা, \fn_cm 4\times 5^{5}\times \frac{1}{4\times 5^{2}}=5^{x}

বা, 5^{3}=5^{x}

{\color{DarkGreen} \therefore x=3}

 

11. সংক্ষিপ্ত উত্তরভিত্তিক প্রশ্নঃ 

(iii) \fn_cm {\color{Blue} 3\times 27^{x}=9^{x+4}} হলে, \fn_cm {\color{Blue} x} -এর মান কত হিসাব করে লিখি।

সমাধানঃ 

\fn_cm 3\times 27^{x}=9^{x+4}

বা, \fn_cm 3\times \left ( 3^{3} \right )^{x}=\left ( 3^{2} \right )^{x+4}

বা, \fn_cm 3\times \left ( 3 \right )^{3x}=\left ( 3 \right )^{2x+8}

বা, \fn_cm \left ( 3 \right )^{3x+1}=\left ( 3 \right )^{2x+8}

বা, 3x+1=2x+8

বা, 3x-2x=8-1

{\color{DarkGreen} \therefore x=7}

 

11. সংক্ষিপ্ত উত্তরভিত্তিক প্রশ্নঃ 

(iv) \fn_cm {\color{Blue} \sqrt[3]{\left ( \frac{1}{64} \right )^{\frac{1}{2}}}} -এর মান কত হিসাব করে লিখি।

সমাধানঃ 

\fn_cm \sqrt[3]{\left ( \frac{1}{64} \right )^{\frac{1}{2}}}

\fn_cm =\sqrt[3]{\left ( \frac{1}{4^{3}} \right )^{\frac{1}{2}}}

\fn_cm ={\left ( \frac{1}{4} \right )^{3\times \frac{1}{2}\times \frac{1}{3}}}

\fn_cm ={\left ( \frac{1}{2^{2}} \right )^{\frac{1}{2}}}

\fn_cm ={\left ( \frac{1}{2} \right )^{2\times \frac{1}{2}}}

\fn_cm =\frac{1}{2}

নির্ণেয়  \fn_cm {\color{DarkGreen} \sqrt[3]{\left ( \frac{1}{64} \right )^{\frac{1}{2}}}} -এর মান {\color{DarkGreen} \frac{1}{2}}

 

11. সংক্ষিপ্ত উত্তরভিত্তিক প্রশ্নঃ 

(v) \fn_cm {\color{Blue} 3^{3^{3}}} এবং \fn_cm {\color{Blue} \left ( 3^{3} \right )^{3}} -এর মধ্যে কোনটি বৃহত্তর যুক্তিসহ লিখি।

সমাধানঃ 

\fn_cm 3^{3^{3}}=3^{27} এবং

\fn_cm \left ( 3^{3} \right )^{3}=3^{9}

নির্ণেয় \fn_cm {\color{DarkGreen} 3^{3^{3}}} এবং \fn_cm {\color{DarkGreen} \left ( 3^{3} \right )^{3}} -এর মধ্যে \fn_cm {\color{DarkGreen} 3^{3^{3}}} সংখ্যাটি বৃহত্তর

 

Koshe dekhi 2 Class 9

Support Me

If you like my work then you can Support me by contributing a small amount which will help me a lot to grow my Website. It’s a request to all of you. You can donate me through phone pay / Paytm/ Gpay  on this number 7980608289 or by the link below :

Subscribe my Youtube channel : Science Duniya in Bangla

and    Learning Science

and visit Our website : learningscience.co.in 

গণিত প্রকাশ (দশম শ্রেণী) সম্পূর্ণ সমাধান

গণিত প্রকাশ (নবম শ্রেণী) সম্পূর্ণ সমাধান

গণিত প্রভা (ষষ্ঠ শ্রেণী) সম্পূর্ণ সমাধান

জীবন বিজ্ঞান  (দশম শ্রেণী) (Life Science)

Thank You

Koshe Dekhi 2 class 9,Koshe Dekhi 2 class 9,Koshe Dekhi 2 class 9

3 thoughts on “Koshe dekhi 2 class 9”

Leave a Reply

Your email address will not be published. Required fields are marked *

Insert math as
Block
Inline
Additional settings
Formula color
Text color
#333333
Type math using LaTeX
Preview
\({}\)
Nothing to preview
Insert
error: Content is protected !!