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

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

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

simplify

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

- x - 1

Solution steps

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

Apply the fraction rule:

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

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

Cancel

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

= - x - 1

\frac{1.1}{1.1 + 1}

(x - 4 y)^{2}

\frac{1}{4 - x}

\frac{6}{40}

\frac{3 ab + 3 a}{b ^{3} + b ^{2}}

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