Question
Prove that: $\sqrt{\frac{1}{4}}+(0.01)^{-\frac{1}{2}}-(27)^{\frac{2}{3}}=\frac{3}{2}$
✓
Answer
We have, $\sqrt{\frac{1}{4}}+(0.01)^{-\frac{1}{2}}-(27)^{\frac{2}{3}}=\frac{3}{2}$
$=\frac{1}{2}+\frac{1}{(0.001)^{\frac{1}{2}}}-(3^3)^{\frac{2}{3}}$
$=\frac{1}{2}+\frac{1}{(0.1)^{2\times\frac{1}{2}}}-3^{3\times\frac{2}{3}}$
$=\frac{1}{2}+\frac{1}{0.1}-3^2$
$=\frac{1}{2}+10-9$
$=\frac{1}{2}+1=\frac{3}{2}$
$\Rightarrow\sqrt{\frac{1}{4}}(0.001)^{-\frac{1}{2}}-(27)^{\frac{2}{3}}=\frac{3}{2}$
$=\frac{1}{2}+\frac{1}{(0.001)^{\frac{1}{2}}}-(3^3)^{\frac{2}{3}}$
$=\frac{1}{2}+\frac{1}{(0.1)^{2\times\frac{1}{2}}}-3^{3\times\frac{2}{3}}$
$=\frac{1}{2}+\frac{1}{0.1}-3^2$
$=\frac{1}{2}+10-9$
$=\frac{1}{2}+1=\frac{3}{2}$
$\Rightarrow\sqrt{\frac{1}{4}}(0.001)^{-\frac{1}{2}}-(27)^{\frac{2}{3}}=\frac{3}{2}$
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