What is the answer to x^{-n-1}= 1/(x^{n+1)} ?

The answer to x^{-n-1}= 1/(x^{n+1)} is True for all x;\quad x^{n+1}\ne 0

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x^{-n-1}= 1/(x^{n+1)}

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

True for all x; x^{n + 1} \neq = 0

Solution steps

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

Multiply both sides by

x^{n + 1}

1 = 1; x^{n + 1} \neq = 0

Both sides are equal

True for all x; x^{n + 1} \neq = 0

a, x = \frac{b}{2 a}

\frac{x ^{2} - 4}{x ^{2} + 3 x + 2} = 0

x, z = \frac{x + y}{m x}

\frac{x}{x - 6} + 3 = \frac{6}{x - 6}

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

What is the answer to x^{-n-1}= 1/(x^{n+1)} ?
The answer to x^{-n-1}= 1/(x^{n+1)} is True for all x;\quad x^{n+1}\ne 0