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

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

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

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

x < - 1 or 0 < x \leq \frac{1}{2} or 1 < x < 2

Interval Notation

(- \infty, - 1) \cup (0, \frac{1}{2}] \cup (1, 2)

Decimal Notation

x < - 1 or 0 < x \leq 0.5 or 1 < x < 2

Solution steps

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

Rewrite in standard form

\frac{4 x - 2}{x ( x + 1 ) ( x - 2 ) ( x - 1 )} \leq 0

Factor

\frac{4 x - 2}{x ( x + 1 ) ( x - 2 ) ( x - 1 )} : \frac{2 ( 2 x - 1 )}{x ( x + 1 ) ( x - 2 ) ( x - 1 )}

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

Divide both sides by

2

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

Simplify

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

Identify the intervals

x < - 1 or 0 < x \leq \frac{1}{2} or 1 < x < 2

\frac{( x ^{2} + 6 )}{( 5 x )} \geq 1

\frac{4}{( ( x - 2 ) ^{2} )} > 1

(x^{2} - x) (x^{3} - x) (x^{4} - x) < 0

\frac{( x + 1 )}{( x ^{2} + 2 )} > \frac{1}{4}

x, f > 0

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