Sat. Jun 22nd, 2024

Jan 27, 2021

Koshe Dekhi 2Class 9

Koshe Dekhi 2 Class 9

1. মান নির্ণয় করি :

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

সমাধানঃ

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

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

$=2^{\frac{3}{5}\times&space;\frac{5}{2}}\times&space;\frac{1}{2^{4\times&space;\frac{3}{2}}}$

$=2^{\frac{3}{2}}\times&space;\frac{1}{2^{6}}$

$=2^{\frac{3}{2}}\times&space;2^{-6}$

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

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

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

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

1. মান নির্ণয় করি :

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

সমাধানঃ

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

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

$=\left&space;\{&space;5^{-6}\times&space;2^{-6}&space;\right&space;\}^{\frac{-1}{6}}$

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

= 5 × 2

= 10

1. মান নির্ণয় করি :

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

সমাধানঃ

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

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

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

$\fn_cm&space;=2^{\frac{2}{3}}\times&space;\left&space;[&space;2^{\frac{1}{3}}\times&space;3^{\frac{1}{2}}&space;\right&space;]\times&space;\frac{1}{3^{\frac{1}{2}}}$

$\fn_cm&space;=2^{\frac{2}{3}}\times&space;2^{\frac{1}{3}}\times&space;3^{\frac{1}{2}}&space;\times&space;3^{-\frac{1}{2}}$

$\fn_cm&space;=2^{\frac{2}{3}+{\frac{1}{3}}}\times&space;3^{\frac{1}{2}-\frac{1}{2}}$

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

$=2^{1}\times&space;3^{0}$

= 2 × 1

= 2

2. সরল করি :

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

সমাধানঃ

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

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

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

$=4a^{2}\times&space;\frac{1}{\frac{1}{9x^{2}}}\times&space;\frac{1}{16a^{2}}\times&space;\frac{1}{9x^{2}}$

$=\frac{1}{4}\times&space;1\times&space;{\frac{9x^{2}}{1}}\times&space;\frac{1}{9x^{2}}$

$=\frac{1}{4}$

2. সরল করি :

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

সমাধানঃ

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

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

$=\left&space;\{&space;\left&space;(&space;\frac{1}{x}&space;\right&space;)^{\frac{10}{3}}&space;\right&space;\}^{\frac{-3}{10}}$

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

$=\left&space;(&space;\frac{1}{x}&space;\right&space;)^{-1}$

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

$=1\times&space;\frac{x}{1}$

$=x$

2. সরল করি :

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

সমাধানঃ

$\fn_cm&space;\left&space;[&space;\left&space;\{&space;\left&space;(&space;2^{-1}&space;\right&space;)^{-1}&space;\right&space;\}^{-1}&space;\right&space;]^{-1}$

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

$\fn_cm&space;=\left&space;[&space;\left&space;\{&space;2&space;\right&space;\}^{-1}&space;\right&space;]^{-1}$

$\fn_cm&space;=\left&space;[&space;\frac{1}{2}\right&space;]^{-1}$

= 2

2. সরল করি :

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

সমাধানঃ

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

$=a^{\frac{-2}{3}}b\times&space;b^{\frac{-2}{3}}c\times&space;c^{\frac{-2}{3}}a$

$=a^{\frac{-2}{3}+1}\times&space;b^{\frac{-2}{3}+1}\times&space;c^{\frac{-2}{3}+1}$

$=a^{\frac{1}{3}}\times&space;b^{\frac{1}{3}}\times&space;c^{\frac{1}{3}}$

$=\left&space;(abc&space;\right&space;)^{\frac{1}{3}}$

$=\sqrt[3]{abc}$

2. সরল করি :

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

সমাধানঃ

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

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

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

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

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

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

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

$=\left&space;(2^{\frac{6m}{2}}\right&space;)^{\frac{1}{m}}$

$=\left&space;(2^{3m}&space;\right&space;)^{\frac{1}{m}}$

$\fn_cm&space;=2^{3}$

= 8

2. সরল করি :

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

সমাধানঃ

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

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

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

$=3^{-6}\times&space;2\times&space;2^{2}\times&space;3^{2}\times&space;3^{4}$

$=3^{-6+6}\times&space;2^{1+2}$

$=3^{0}\times&space;2^{3}$

= 1 × 8

= 8

2. সরল করি :

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

সমাধানঃ

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

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

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

$=x^{0}$

= 1

3. মানের ঊর্ধ্বক্রমানুসারে সাজাই :

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

সমাধানঃ

$\fn_cm&space;5^{\frac{1}{2}},10^{\frac{1}{4}},6^{\frac{1}{3}}$

উপরের সংখ্যাগুলির সূচকগুলি হলো  $\fn_cm&space;\frac{1}{2},\frac{1}{4}$$\fn_cm&space;\frac{1}{3}$

2, 4 ও 3 -এর ল.সা.গু. = 12

$\fn_cm&space;\frac{1}{2}=\frac{1\times&space;6}{2\times&space;6}=\frac{6}{12}$

$\fn_cm&space;\frac{1}{4}=\frac{1\times&space;3}{4\times&space;3}=\frac{3}{12}$

$\fn_cm&space;\frac{1}{3}=\frac{1\times&space;4}{3\times&space;4}=\frac{4}{12}$

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

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

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

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

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

3. মানের ঊর্ধ্বক্রমানুসারে সাজাই :

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

সমাধানঃ

$\fn_cm&space;3^{\frac{1}{3}},2^{\frac{1}{2}},8^{\frac{1}{4}}$

উপরের সংখ্যাগুলির সূচকগুলি হলো   $\fn_cm&space;\frac{1}{3}$ , $\fn_cm&space;\frac{1}{2}$$\frac{1}{4}$

3, 2 ও 4 -এর ল.সা.গু. = 12

$\fn_cm&space;\frac{1}{3}=\frac{1\times&space;4}{3\times&space;4}=\frac{4}{12}$

$\fn_cm&space;\frac{1}{2}=\frac{1\times&space;6}{2\times&space;6}=\frac{6}{12}$

$\fn_cm&space;\frac{1}{4}=\frac{1\times&space;3}{4\times&space;3}=\frac{3}{12}$

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

$\fn_cm&space;2^{\frac{1}{2}}=2^{\frac{6}{12}}=\sqrt[12]{2^{6}}=\sqrt[12]{64}$

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

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

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

3. মানের ঊর্ধ্বক্রমানুসারে সাজাই :

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

সমাধানঃ

$\fn_cm&space;2^{60},3^{48},4^{36},5^{24}$

উপরের সংখ্যাগুলির সূচকগুলি হলো   60 , 48 , 36 ও 24

60, 48, 36 ও 24 -এর গ.সা.গু. = 12

$\fn_cm&space;\frac{1}{3}=\frac{1\times&space;4}{3\times&space;4}=\frac{4}{12}$

$\fn_cm&space;\frac{1}{2}=\frac{1\times&space;6}{2\times&space;6}=\frac{6}{12}$

$\fn_cm&space;\frac{1}{4}=\frac{1\times&space;3}{4\times&space;3}=\frac{3}{12}$

এখন, $\fn_cm&space;2^{60}=2^{\frac{60}{12}}=2^{5}=32$

$\fn_cm&space;3^{48}=3^{\frac{48}{12}}=3^{4}=81$

$\fn_cm&space;4^{36}=4^{\frac{36}{12}}=4^{3}=48$

$\fn_cm&space;5^{24}=5^{\frac{24}{12}}=5^{2}=25$

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) $\fn_cm&space;{\color{Blue}&space;\left&space;(&space;\frac{a^{q}}{a^{r}}&space;\right&space;)^{p}\times&space;\left&space;(&space;\frac{a^{r}}{a^{p}}&space;\right&space;)^{q}\times&space;\left&space;(&space;\frac{a^{p}}{a^{q}}&space;\right&space;)^{r}=1}$

সমাধানঃ

বামপক্ষ :

$\fn_cm&space;\left&space;(&space;\frac{a^{q}}{a^{r}}&space;\right&space;)^{p}\times&space;\left&space;(&space;\frac{a^{r}}{a^{p}}&space;\right&space;)^{q}\times&space;\left&space;(&space;\frac{a^{p}}{a^{q}}&space;\right&space;)^{r}$

$=\frac{a^{pq}}{a^{pr}}\times&space;\frac{a^{qr}}{a^{pq}}\times&space;\frac{a^{pr}}{a^{qr}}$

= 1

= ডানপক্ষ (প্রমাণিত)

4. প্রমান করি :

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

সমাধানঃ

বামপক্ষ :

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

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

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

$=x^{0}$

= 1

= ডানপক্ষ (প্রমাণিত)

4. প্রমান করি :

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

সমাধানঃ

বামপক্ষ :

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

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

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

$=x^{0}$

= 1

= ডানপক্ষ (প্রমাণিত)

4. প্রমান করি :

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

সমাধানঃ

বামপক্ষ :

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

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

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

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

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

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

$=a^{0}$

= 1

= ডানপক্ষ (প্রমাণিত)

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

সমাধানঃ

প্রদত্ত,

x + z = 2y

বা, x – y = y – z

এবং,

$\fn_cm&space;b^{2}=ac$

বা, $b=\sqrt{ac}$

বামপক্ষ :

$\fn_cm&space;a^{y-z}b^{z-x}c^{x-y}$

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

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

$=a^{x-y}\times&space;c^{x-y}\times&space;a^{\frac{z-x}{2}}\times&space;c^{\frac{z-x}{2}}$

$=a^{x-y+\frac{z-x}{2}}\times&space;c^{x-y+\frac{z-x}{2}}$

$=a^{\frac{2x-2y+z-x}{2}}\times&space;c^{\frac{2x-2y+z-x}{2}}$

$=a^{\frac{x+z-2y}{2}}\times&space;c^{\frac{x+z-2y}{2}}$

$=a^{\frac{2y-2y}{2}}\times&space;c^{\frac{2y-2y}{2}}$

$=a^{\frac{0}{2}}\times&space;c^{\frac{0}{2}}$

$=a^{0}\times&space;c^{0}$

= 1 × 1

= 1

= ডানপক্ষ (প্রমাণিত)

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

সমাধানঃ

বামপক্ষ :

$\fn_cm&space;a^{q-r}b^{r-p}c^{p-q}$

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

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

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

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

$=x^{0}\times&space;y^{0}$

= 1 × 1

= 1

= ডানপক্ষ (প্রমাণিত)

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

সমাধানঃ

$\fn_cm&space;x^{\frac{1}{a}}=y^{\frac{1}{b}}=z^{\frac{1}{c}}$

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

$\therefore&space;x=k^{a},y=k^{b},z=k^{c}$

এখন,

$xyz=1$

বা, $k^{a}\times&space;k^{b}\times&space;k^{c}=1$

বা, $k^{a+b+c}=1$

বা, $k^{a+b+c}=k^{0}$

$\therefore&space;a+b+c=0$ (প্রমানিত)

Koshe dekhi 2 Class 9

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

সমাধানঃ

$\fn_cm&space;x^{a}=b^{y}=z^{c}$

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

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

আবার,

$abc=1$

এখন,

$\fn_cm&space;xy+yz+zx$

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

বা, $k^{a}\times&space;k^{b}\times&space;k^{c}=1$

বা, $k^{a+b+c}=1$

বা, $k^{a+b+c}=k^{0}$

$\therefore&space;a+b+c=0$ (প্রমানিত)

9. সমাধান করি

(i) $\fn_cm&space;{\color{Blue}&space;49^{x}=7^{3}}$

সমাধানঃ

$\fn_cm&space;49^{x}=7^{3}$

বা, $\left&space;(7^{2}&space;\right&space;)^{x}=7^{3}$

বা, $7^{2x}=7^{3}$

বা, $2x=3$

$\therefore&space;x=\frac{3}{2}$

নির্ণেয় x এর মান ${\color{DarkGreen}&space;\frac{3}{2}}$

9. সমাধান করি

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

সমাধানঃ

$\fn_cm&space;2^{x+2}+2^{x-1}=9$

বা, $\fn_cm&space;2^{x}\times&space;2^{2}+2^{x}\times&space;2^{-1}=9$

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

বা, $\fn_cm&space;2^{x}\times&space;\frac{9}{2}=9$

বা, $\fn_cm&space;2^{x}=9\times&space;\frac{2}{9}$

বা, $\fn_cm&space;2^{x}=2^{1}$

$\therefore&space;x=1$

নির্ণেয় x এর মান 1

9. সমাধান করি

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

সমাধানঃ

$\fn_cm&space;2^{x+1}+2^{x+2}=48$

বা, $\fn_cm&space;2^{x}\times&space;2^{1}+2^{x}\times&space;2^{2}=48$

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

বা, $\fn_cm&space;2^{x}\times&space;6=48$

বা, $\fn_cm&space;2^{x}=48\times&space;\frac{1}{6}$

বা, $\fn_cm&space;2^{x}=8$

বা, $\fn_cm&space;2^{x}=2^{3}$

$\therefore&space;x=3$

নির্ণেয় x এর মান 3

9. সমাধান করি

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

সমাধানঃ

$\fn_cm&space;2^{4x}.4^{3x-1}=\frac{4^{2x}}{2^{3x}}$

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

বা, $\fn_cm&space;2^{4x}\times&space;2^{6x-2}=2^{4x-3x}$

বা, $2^{4x+6x-2}=2^{x}$

বা, $2^{10x-2}=2^{x}$

বা, $2^{10x}\times&space;2^{-2}=2^{x}$

বা, $\frac{2^{10x}}{4}=2^{x}$

বা, $\frac{2^{10x}}{2^{x}}=2^{2}$

বা, $2^{10x-x}=2^{2}$

বা, $9x=2$

$\therefore&space;x=\frac{2}{9}$

নির্ণেয় x এর মান $\fn_cm&space;{\color{DarkGreen}&space;\frac{2}{9}}$

9. সমাধান করি

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

সমাধানঃ

$\fn_cm&space;9\times&space;81^{x}=27^{2-x}$

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

বা, $\fn_cm&space;3^{2}\times&space;3^{4x}=3^{6-3x}$

বা, $\fn_cm&space;3^{2+4x}=3^{6-3x}$

বা, $\fn_cm&space;2+4x=6-3x$

বা, $4x+3x=6-2$

বা, $7x=4$

$\therefore&space;x=\frac{4}{7}$

নির্ণেয় x এর মান ${\color{DarkGreen}&space;\frac{4}{7}}$

9. সমাধান করি

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

সমাধানঃ

$\fn_cm&space;2^{5x+4}+2^{9}=2^{10}$

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

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

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

বা, $\fn_cm&space;2^{5x}=\frac{2^{9}}{2^{4}}$

বা, $\fn_cm&space;2^{5x}=2^{5}$

বা, $\fn_cm&space;5x=5$

${\color{DarkGreen}&space;\therefore&space;x=\frac{5}{5}=1}$

9. সমাধান করি

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

সমাধানঃ

$\fn_cm&space;6^{2x+4}=3^{3x}.2^{x+8}$

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

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

বা, $\fn_cm&space;\frac{3^{2x+4}}{3^{3x}}=\frac{2^{x+8}}{2^{2x+4}}$

বা, $\fn_cm&space;3^{2x+4-3x}=2^{x+8-2x-4}$

বা, $\fn_cm&space;3^{4-x}=2^{4-x}$

বা, $\fn_cm&space;\frac{3^{4}}{3^{x}}=\frac{2^{4}}{2^{x}}$

বা, $\fn_cm&space;\frac{3^{4}}{2^{4}}=\frac{3^{x}}{2^{x}}$

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

${\color{DarkGreen}&space;\therefore&space;x=4}$

Koshe dekhi 2 Class 9

10. বহু বিকল্পীয় প্রশ্ন (M.C.Q.) :

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

(a) 0.3

(b) 3

(c) 0.9

(d) 9

সমাধানঃ

$\fn_cm&space;\left&space;(&space;0.243&space;\right&space;)^{0.2}\times&space;\left&space;(&space;10&space;\right&space;)^{0.6}$

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

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

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

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

$=3$

উত্তরঃ (b) 3

10. বহু বিকল্পীয় প্রশ্ন (M.C.Q.) :

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

(a) 1

(b) 2

(c) 4

(d) $\fn_cm&space;{\color{Blue}&space;\frac{1}{2}}$

সমাধানঃ

$\fn_cm&space;2^{\frac{1}{2}}\times&space;2^{-\frac{1}{2}}\times&space;\left&space;(&space;16&space;\right&space;)^{\frac{1}{2}}$

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

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

$\fn_cm&space;=2^{\frac{1}{2}-\frac{1}{2}+2}$

$=2^{2}$

= 4

উত্তরঃ (c) 4

10. বহু বিকল্পীয় প্রশ্ন (M.C.Q.) :

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

(a) $\fn_cm&space;{\color{Blue}&space;\frac{3}{2}}$

(b) $\fn_cm&space;{\color{Blue}&space;\frac{9}{2}}$

(c) 3

(d) 9

সমাধানঃ

$\fn_cm&space;4^{x}=8^{3}$

বা, $\left&space;(&space;2^{2}&space;\right&space;)^{x}=\left&space;(&space;2^{3}&space;\right&space;)^{3}$

বা, $2^{2x}=2^{9}$

বা, $2x=9$

$\therefore&space;x=\frac{9}{2}$

উত্তরঃ (b) $\fn_cm&space;{\color{DarkGreen}&space;\frac{9}{2}}$

10. বহু বিকল্পীয় প্রশ্ন (M.C.Q.) :

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

(a) $\fn_cm&space;{\color{Blue}&space;\frac{1}{49}}$

(b) 7

(c) 49

(d) 1

সমাধানঃ

$\fn_cm&space;20^{-x}=\frac{1}{7}$

বা, $\frac{1}{20^{x}}=\frac{1}{7}$

বা, $20^{x}=7$

বা, $\left&space;(20^{x}&space;\right&space;)^{2}=7^{2}$

$\therefore&space;20^{2x}=49$

উত্তরঃ (c) 49

10. বহু বিকল্পীয় প্রশ্ন (M.C.Q.) :

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

(a) 8

(b) 1

(c) 64

(d) 27

সমাধানঃ

$\fn_cm&space;4\times&space;5^{x}=500$

বা, $\fn_cm&space;4\times&space;5^{x}=4\times&space;5^{3}$

বা, $\fn_cm&space;5^{x}=5^{3}$

$\therefore&space;x=3$

$\fn_cm&space;\therefore&space;x^{x}$ -এর মান $=3^{3}=27$

উত্তরঃ (d) 27

Koshe dekhi 2 Class 9

11. সংক্ষিপ্ত উত্তরভিত্তিক প্রশ্নঃ

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

সমাধানঃ

$\fn_cm&space;\left&space;(&space;27&space;\right&space;)^{x}=\left&space;(&space;81&space;\right&space;)^{y}$

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

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

বা, $3x=4y$

বা, $\frac{x}{y}=\frac{4}{3}$

${\color{DarkGreen}&space;\therefore&space;x:y=4:3}$

11. সংক্ষিপ্ত উত্তরভিত্তিক প্রশ্নঃ

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

সমাধানঃ

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

ধরি, ${\color{Purple}&space;5^{5}=a,0.01=b}$

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

বা, $\fn_cm&space;4\times&space;5^{5}\times&space;\frac{1}{100}=5^{x}$

বা, $\fn_cm&space;4\times&space;5^{5}\times&space;\frac{1}{4\times&space;5^{2}}=5^{x}$

বা, $5^{3}=5^{x}$

${\color{DarkGreen}&space;\therefore&space;x=3}$

11. সংক্ষিপ্ত উত্তরভিত্তিক প্রশ্নঃ

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

সমাধানঃ

$\fn_cm&space;3\times&space;27^{x}=9^{x+4}$

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

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

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

বা, $3x+1=2x+8$

বা, $3x-2x=8-1$

${\color{DarkGreen}&space;\therefore&space;x=7}$

11. সংক্ষিপ্ত উত্তরভিত্তিক প্রশ্নঃ

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

সমাধানঃ

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

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

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

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

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

$\fn_cm&space;=\frac{1}{2}$

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

11. সংক্ষিপ্ত উত্তরভিত্তিক প্রশ্নঃ

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

সমাধানঃ

$\fn_cm&space;3^{3^{3}}=3^{27}$ এবং

$\fn_cm&space;\left&space;(&space;3^{3}&space;\right&space;)^{3}=3^{9}$

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

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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”
1. Asmita says:

Thanks .

This is very simple type

Thank you for your valuable comment.

1. Asmita says:

Thanks for this

Koshe dekhi 20.1 Class 8

Insert math as
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