Question
Find the value $ \left[4^{-1}+3^{-1}+6^{-2}\right]^{-1}$
✓
Answer
Using law of exponents, $\text{a}^{-\text{m}}=\frac{1}{\text{a}^\text{m}}[ \because$ a is non-zero integer$]$
$\therefore\left[4^{-1}+3^{-1}+6^{-2}\right]^{-1}$
$=\Big(\frac{1}{4}+\frac{1}{3}+\frac{1}{36}\Big)^{-1}$
$=\Big(\frac{9+12+1}{36}\Big)^{-1}[\because LCM$ of $4, 3,$ and $36 = 36]$
$=\Big(\frac{22}{36}\Big)^{-1}=\frac{36}{22}=\frac{18}{11}$
Hence,
$\Big[4^{-1}+3^{-1}+6^{-2}\Big]^{-1}=\frac{18}{11}$
$\therefore\left[4^{-1}+3^{-1}+6^{-2}\right]^{-1}$
$=\Big(\frac{1}{4}+\frac{1}{3}+\frac{1}{36}\Big)^{-1}$
$=\Big(\frac{9+12+1}{36}\Big)^{-1}[\because LCM$ of $4, 3,$ and $36 = 36]$
$=\Big(\frac{22}{36}\Big)^{-1}=\frac{36}{22}=\frac{18}{11}$
Hence,
$\Big[4^{-1}+3^{-1}+6^{-2}\Big]^{-1}=\frac{18}{11}$
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