Prove that:$ \frac{(0.6)^{0}-(0.1)^{-1}}{\left(\frac{3}{8}\right)^{-1}\left(\frac{3}{2}\right)^{3}+\left(-\frac{1}{3}\right)^{-1}}=-\frac{3}{2} $

Given: 

\( \frac{(0.6)^{0}-(0.1)^{-1}}{\left(\frac{3}{8}\right)^{-1}\left(\frac{3}{2}\right)^{3}+\left(-\frac{1}{3}\right)^{-1}}=-\frac{3}{2} \)

To do: 

We have to prove that \( \frac{(0.6)^{0}-(0.1)^{-1}}{\left(\frac{3}{8}\right)^{-1}\left(\frac{3}{2}\right)^{3}+\left(-\frac{1}{3}\right)^{-1}}=-\frac{3}{2} \).

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 $=\frac{(0.6)^{0}-(0.1)^{-1}}{(\frac{3}{8})^{-1} \times(\frac{3}{2})^{3}+(-\frac{1}{3})^{-1}}$

$=\frac{1-(\frac{1}{10})^{-1}}{(\frac{8}{3})^{1} \times(\frac{3}{2})^{3}+(\frac{-3}{1})^{1}}$

$=\frac{1-10^{1}}{\frac{8}{3} \times \frac{27}{8}+(\frac{-3}{1})}$

$=\frac{-9}{9-\frac{3}{1}}$

$=\frac{-9}{6}$

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

$=$ RHS

Hence proved.      

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Updated on: 10-Oct-2022

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