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

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

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

simplify

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

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

Solution steps

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

Apply the fraction rule:

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

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

Cancel

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

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

(- 1.3 p + 6.2) (2.6)

3 p^{\frac{1}{2}} \cdot p^{3}

(\frac{( - 5 - 3 b )}{6}) - (\frac{( b - 4 - a )}{2})

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

\frac{5 n ^{4} + 30 n ^{3}}{n ^{2}}

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