Prove that:$ \sqrt{\frac{1}{4}}+(0.01)^{-1 / 2}-(27)^{-2 / 3}=\frac{3}{2} $

Given: 

\( \sqrt{\frac{1}{4}}+(0.01)^{-1 / 2}-(27)^{-2 / 3}=\frac{3}{2} \)

To do: 

We have to prove that \( \sqrt{\frac{1}{4}}+(0.01)^{-1 / 2}-(27)^{-2 / 3}=\frac{3}{2} \).

Solution:

We know that,

$(a^{m})^{n}=a^{m n}$

$a^{m} \times a^{n}=a^{m+n}$

$a^{m} \div a^{n}=a^{m-n}$

$a^{0}=1$

Therefore,

LHS $=\sqrt{\frac{1}{4}}+(0.01)^{\frac{-1}{2}}-(27)^{\frac{2}{3}}$

$=(\frac{1}{2^{2}})^{\frac{1}{2}}+(0.1)^{2 \times(\frac{-1}{2})}-(3^{3})^{\frac{2}{3}}$

$=\frac{1}{(2^{2})^{\frac{1}{2}}}+(0.1)^{2 \times(\frac{-1}{2})}-3^{3 \times \frac{2}{3}}$

$=\frac{1}{2^{1}}+(0.1)^{-1}-3^{2}$

$=\frac{1}{2}+(\frac{1}{10})^{-1}-3^{2}$

$=\frac{1}{2}+10-9$

$=\frac{1}{2}+1$

$=\frac{3}{2}$

$=$ RHS

Hence proved.   

Tutorialspoint

Updated on: 10-Oct-2022

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