1<|(x-2)/(x+3)|

解答

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

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

求解步骤

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

交换两边

\frac{x - 2}{x + 3} > 1

使用绝对值运算法则: 若

∣ u ∣ > a, a > 0

，则

u < - a

u > a

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

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

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

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

\frac{3 !}{0 ! ( 3 - 0 ) !}

3 z^{2} + 19 z + 20

\frac{2}{3} x - 5 = 7

2 x - \frac{16}{5} \leq \frac{1}{3}