What is 1/(x+1)+1/(x+2)-(4x)/(x^2+3x+2)<0 ?

The solution to 1/(x+1)+1/(x+2)-(4x)/(x^2+3x+2) 3/2

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1/(x+1)+1/(x+2)-(4x)/(x^2+3x+2)<0

\frac{1}{x + 1} + \frac{1}{x + 2} - \frac{4 x}{x ^{2} + 3 x + 2} < 0

- 2 < x < - 1 or x > \frac{3}{2}

Interval Notation

(- 2, - 1) \cup (\frac{3}{2}, \infty)

Decimal Notation

- 2 < x < - 1 or x > 1.5

Solution steps

\frac{1}{x + 1} + \frac{1}{x + 2} - \frac{4 x}{x ^{2} + 3 x + 2} < 0

Rewrite in standard form

\frac{- 2 x + 3}{( x + 1 ) ( x + 2 )} < 0

Identify the intervals

- 2 < x < - 1 or x > \frac{3}{2}

2 x^{2} + 12 x + 8 = 8 x + 4

\frac{- 5 + ( 5 ) ^{2} - 4 ( 3 ) ( - 2 )}{2 ( 3 )} = x

3 x + 2 < 2 x - 4

2 r^{2} + 5 r - 3 = 0

(7 x - 3 y) (2 x - 3 y) - (3 y - 2 x)^{2}

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