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

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

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1/x <= 1/(x+1)-2/(x+2)

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

x < - 2 or - 1 < x < 0

Interval Notation

(- \infty, - 2) \cup (- 1, 0)

Solution steps

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

Rewrite in standard form

\frac{2 x ^{2} + 3 x + 2}{x ( x + 2 ) ( x + 1 )} \leq 0

Identify the intervals

x < - 2 or - 1 < x < 0

∣ x - 3 ∣ < x + 2

2 n^{4} + 8 n^{3} + 11 n^{2} + 6 n + 1

Y = 0.00007 \cdot 1. 2^{2} - 0.003 \cdot 1.2 + 0.0063

0 + (9.76 - \frac{12.5}{103} (0)) 0.5

x^{2} - x - 5 < 3

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