Question

নির্ণয়েয় সমাধান৷৷ ???????

Answer 1

[ Answer = 8 ]

Solution :

$\displaystyle \mathrm{Now,\:\bigg(\frac{4^{m+\frac{1}{4}}\times \sqrt{2.2^{m}}}{2.\sqrt{2^{-m}}}\bigg)^{\frac{1}{m}}}$

$\displaystyle \mathrm{=\bigg(\frac{(2^{2})^{m+\frac{1}{4}}.(2^{m+1})^{\frac{1}{2}}}{2.2^{-\frac{m}{2}}}\bigg)^{\frac{1}{m}}}$

$\displaystyle \mathrm{=\bigg(\frac{2^{2m+\frac{1}{2}}.2^{\frac{m}{2}+\frac{1}{2}}}{2^{1-\frac{m}{2}}}\bigg)^{\frac{1}{m}}}$

$\displaystyle \mathrm{=\bigg(2^{2m+\frac{1}{2}+\frac{m}{2}+\frac{1}{2}-1+\frac{m}{2}}\bigg)^{\frac{1}{m}}}$

$\displaystyle \mathrm{=(2^{3m})^{\frac{1}{m}}}$

$\displaystyle \mathrm{=2^{\frac{3m}{m}}}$

$\displaystyle \mathrm{=2^{3}}$

= 8

$\displaystyle \implies \boxed{\mathrm{\bigg(\frac{4^{m+\frac{1}{4}}\times \sqrt{2.2^{m}}}{2.\sqrt{2^{-m}}}\bigg)^{\frac{1}{m}}=8}}$