What is (\frac{16-x^2)/(x^2+2x)}{(x-4)/(x^2-4)} ?

The solution to (\frac{16-x^2)/(x^2+2x)}{(x-4)/(x^2-4)} is -((4+x)(x-2))/x

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simplify (\frac{16-x^2)/(x^2+2x)}{(x-4)/(x^2-4)}

simplify

\frac{\frac{16 - x ^{2}}{x ^{2} + 2 x}}{\frac{x - 4}{x ^{2} - 4}}

- \frac{( 4 + x ) ( x - 2 )}{x}

Solution steps

\frac{\frac{16 - x ^{2}}{x ^{2} + 2 x}}{\frac{x - 4}{x ^{2} - 4}}

Apply the fraction rule:

\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a \cdot d}{b \cdot c}

= \frac{( 16 - x ^{2} ) ( x ^{2} - 4 )}{( x ^{2} + 2 x ) ( x - 4 )}

Cancel

\frac{( 16 - x ^{2} ) ( x ^{2} - 4 )}{( x ^{2} + 2 x ) ( x - 4 )} : - \frac{( 4 + x ) ( x - 2 )}{x}

= - \frac{( 4 + x ) ( x - 2 )}{x}

\frac{( 1 + 3 i ) ^{8} ( 3 - i ) ^{6}}{( - 1 + i ) ^{4}}

\frac{( 4 x ^{2} y z ^{3} ) ^{2}}{2 x ^{3} y ^{2} z ^{- 4}}

1^{2} + 1^{2} + 1^{2}

\frac{2}{3} \div \frac{4}{3}

(4 + 6 i) - (8 + 3 i)

What is (\frac{16-x^2)/(x^2+2x)}{(x-4)/(x^2-4)} ?
The solution to (\frac{16-x^2)/(x^2+2x)}{(x-4)/(x^2-4)} is -((4+x)(x-2))/x