((x+5)(x-9))/((x-12))>0

解答

\frac{( x + 5 ) ( x - 9 )}{( x - 12 )} > 0

- 5 < x < 9 or x > 12

间隔符号

(- 5, 9) \cup (12, \infty)

求解步骤

\frac{( x + 5 ) ( x - 9 )}{x - 12} > 0

确定区间

- 5 < x < 9 or x > 12

(x^{2} - 4) (x^{3} - 5 x^{2} + 6 x) \geq 0

x^{3} - 6 x^{2} + 11 x - 6 < 0

\frac{( x + 3 )}{3} - \frac{4}{( x + 2 )} > \frac{x}{3}

\frac{x ^{- 5} x}{x - 2} > 0

\frac{( 5 x + 1 )}{( ( x + 1 ) ^{2} )} < 1