Question

solve it.
Find the value of (1/2)^-2 + (1/3)^-2 + (1/2)^-2​

Answer 1

We have given $\large{\bf{\green{(\frac{1}{2})^{-2} + (\frac{1}{3})^{-2} + (\frac{1}{4})^{-2}}}}$. So to solve these type of questions, we required some formulas of Exponent. In this question, we will use a formula which says that $(\frac{1}{n})^{-m}$ can be written as ${n}^{m}$.

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Let's understand that what does this formula saying:

This formula says that, if power of any number is in negative form and if we will reverse the number, then the power became positive. Similarly, in this question also, we have given 3 numbers (in fraction form) and powers are negative. So for solving further, we need to make power positive. Using this formula (mentioned above), if we reverse the number then power became positive.

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Considering above formula:

$\large{\bf{(\frac{1}{2})^{-2} + (\frac{1}{3})^{-2} + (\frac{1}{4})^{-2}}}$

$\implies\large{\bf{2^2+3^2+4^2}}$

$\implies\large{\bf{4+9+16}}$

$\implies\large{\bf{13+16}}$

$\implies\large{\bf{29}}$

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Know more formulas related to this type of questions:

• $\large{\sf{(\frac{a}{b})^m\times(\frac{a}{b})^n~=~(\frac{a}{b})^{m+n}}}$
• $\large{\sf{\bigg[(\frac{a}{b})^m\bigg]^n~=~(\frac{a}{b})^{mn}}}$
• $\large{\sf{(\frac{a}{b})^m\times(\frac{c}{d})^m~=~\bigg({\frac{a}{b}\times {\frac{c}{d}\bigg)^m}}}}$
• $\large{\sf{(\frac{a}{b})^m\div (\frac{a}{b})^n~=~(\frac{a}{b})^{m-n}}}$

Answer 2

Step-by-step explanation:

$( \frac{1}{2})^{ - 2} + ( \frac{1}{3} )^{ - 2} + ( \frac{1}{4} )^{ - 2}$

$( \frac{2}{1} )^{2} + {( \frac{3}{1} })^{2} + {( \frac{4}{1} })^{2}$

$4 + 9 + 16$

$29$

I hope it is helpful