Gujarat BoardEnglish MediumSTD 8MATHSExponents and Powers1 Mark
Question
The value of $\left[1^{-2}+2^{-2}+3^{-2}\right] \times 6^2$ is ________ .
✓
Answer
The value of $\left[1^{-2}+2^{-2}+3^{-2}\right] \times 6^2$ is $49.$ Solution:
Using law of exponents, $\text{a}^{-\text{m}}=\frac{1}{\text{a}^\text{m}}$ [$\because$ a is non-zero integer]
$\therefore \big[1^{-2}+2^{-2}+3^{-2}\big]\times6^2$
$=\Big[\frac{1}{2^2}+\frac{1}{2^2}+\frac{1}{3^3}\Big]\times6^2$
$=\Big[1+\frac{1}{4}+\frac{1}{9}\Big]\times6^2$
$=\Big(\frac{36+9+4}{36}\Big)\times6^2$
$=\Big(\frac{49}{36}\Big)\times6^2=\Big(\frac{7}{6}\Big)^2\times6^2$
$(7)^2\times6^{-2}\times6^2$
$=(7)^2\times6^{2-2}=(7)^2\times6^0=49$
$[a^0= 1]$
Hence,$ [1^{-2}+ 2^{-2}+ 3^{-2}] \times 6^2= 49$
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