What is (x^{-2}+2^{-2})^{-1} ?

The solution to (x^{-2}+2^{-2})^{-1} is (4x^2)/(4+x^2)

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simplify (x^{-2}+2^{-2})^{-1}

simplify

(x^{- 2} + 2^{- 2})^{- 1}

\frac{4 x ^{2}}{4 + x ^{2}}

Solution steps

(x^{- 2} + 2^{- 2})^{- 1}

Apply exponent rule:

a^{- 1} = \frac{1}{a}

(x^{- 2} + 2^{- 2})^{- 1} = \frac{1}{x ^{- 2} + 2 ^{- 2}}

= \frac{1}{x ^{- 2} + 2 ^{- 2}}

Simplify

x^{- 2} + 2^{- 2} : \frac{1}{x ^{2}} + \frac{1}{2 ^{2}}

= \frac{1}{\frac{1}{x ^{2}} + \frac{1}{2 ^{2}}}

Join

\frac{1}{x ^{2}} + \frac{1}{2 ^{2}} : \frac{4 + x ^{2}}{4 x ^{2}}

= \frac{1}{\frac{4 + x ^{2}}{4 x ^{2}}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{4 x ^{2}}{4 + x ^{2}}

- \frac{1}{6} + \frac{5}{6}

7 p - 1 - 9 p + 5

(\frac{8}{125})^{- \frac{1}{3}}

\frac{18}{56}

What is (x^{-2}+2^{-2})^{-1} ?
The solution to (x^{-2}+2^{-2})^{-1} is (4x^2)/(4+x^2)