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

解答

\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}

间隔符号

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

十进制

- 2 < x < - 1 or x > 1.5

求解步骤

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

改写为标准形式

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

确定区间

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

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

3 x + 2 < 2 x - 4

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

s im pl i f y (7 x - 3 y) (2 x - 3 y) - (3 y - 2 x)^{2}

\frac{∣ x - 1 ∣}{3 x - 1} \geq 0