What is (x+3)/(x^2-4)-1/(x+2)<(2x)/(2x-x^2) ?

The solution to (x+3)/(x^2-4)-1/(x+2)<(2x)/(2x-x^2) is x<-9/2 \lor-2<x<0\lor 0<x<2

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(x+3)/(x^2-4)-1/(x+2)<(2x)/(2x-x^2)

\frac{x + 3}{x ^{2} - 4} - \frac{1}{x + 2} < \frac{2 x}{2 x - x ^{2}}

x < - \frac{9}{2} or - 2 < x < 0 or 0 < x < 2

Interval Notation

(- \infty, - \frac{9}{2}) \cup (- 2, 0) \cup (0, 2)

Decimal Notation

x < - 4.5 or - 2 < x < 0 or 0 < x < 2

Solution steps

\frac{x + 3}{x ^{2} - 4} - \frac{1}{x + 2} < \frac{2 x}{2 x - x ^{2}}

Rewrite in standard form

\frac{2 x ^{2} + 9 x}{x ( x + 2 ) ( x - 2 )} < 0

Factor

\frac{2 x ^{2} + 9 x}{x ( x + 2 ) ( x - 2 )} : \frac{x ( 2 x + 9 )}{x ( x + 2 ) ( x - 2 )}

\frac{x ( 2 x + 9 )}{x ( x + 2 ) ( x - 2 )} < 0

Identify the intervals

x < - \frac{9}{2} or - 2 < x < 0 or 0 < x < 2

3 - 3 x^{2} = 0

24 > 4 w

2 r^{2} - 3 r + 1

\frac{0}{- 4}

\frac{7 - x}{x - 4} > 0

What is (x+3)/(x^2-4)-1/(x+2)<(2x)/(2x-x^2) ?
The solution to (x+3)/(x^2-4)-1/(x+2)<(2x)/(2x-x^2) is x<-9/2 \lor-2<x<0\lor 0<x<2