(3x^2-x-4)(x-1)(2x-3)>0

解答

(3 x^{2} - x - 4) (x - 1) (2 x - 3) > 0

x < - 1 or 1 < x < \frac{4}{3} or x > \frac{3}{2}

间隔符号

(- \infty, - 1) \cup (1, \frac{4}{3}) \cup (\frac{3}{2}, \infty)

十进制

x < - 1 or 1 < x < 1.33333 \dots or x > 1.5

求解步骤

(3 x^{2} - x - 4) (x - 1) (2 x - 3) > 0

分解

(3 x^{2} - x - 4) (x - 1) (2 x - 3) : (x + 1) (x - 1) (3 x - 4) (2 x - 3)

(x + 1) (x - 1) (3 x - 4) (2 x - 3) > 0

确定区间

x < - 1 or 1 < x < \frac{4}{3} or x > \frac{3}{2}

x^{2} + 2 x - 10 > 2 (x - 5)

\frac{( x ^{2} - 1 )}{( x ^{2} )} > 0

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

(2 x^{2} - 13 x - 27) (x^{2} - 2 x + 6) \leq 0

x^{2} - 3 x + 4 > - 2