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

The solution to 1/(x^2+x)>= 1/(x^2-x)-1/(x^2-x) is x 0

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

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

x < - 1 or x > 0

Solution steps

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

Rewrite in standard form

\frac{1}{x ^{2} + x} \geq 0

\frac{1}{a} \geq 0

then

a > 0

x^{2} + x > 0

Factor

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

x (x + 1) > 0

Identify the intervals

x < - 1 or x > 0

350 \leq 5.5 t - 110

12 x - 9 = 6 x + 39

\frac{8}{4}

- 15 = m - 14

x \leq 0 and x \geq - 4

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