What is 1/(x^2+x)> 1/(x^2-x)-1/(x^2-1) ?

The solution to 1/(x^2+x)> 1/(x^2-x)-1/(x^2-1) is x 2

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1/(x^2+x)> 1/(x^2-x)-1/(x^2-1)

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

x < - 1 or 0 < x < 1 or x > 2

Solution steps

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

Rewrite in standard form

\frac{x - 2}{x ( x + 1 ) ( x - 1 )} > 0

Identify the intervals

x < - 1 or 0 < x < 1 or x > 2

lo g_{3} (2 7^{5})

\frac{2 t - 1}{6} = \frac{t + 2}{4} + \frac{1}{3}

700

25 x^{2} - 90 x + 82 = 0

\frac{1}{- 5}

What is 1/(x^2+x)> 1/(x^2-x)-1/(x^2-1) ?
The solution to 1/(x^2+x)> 1/(x^2-x)-1/(x^2-1) is x<-1\lor 0<x<1\lor x>2