(x^2-12x+32)(x-1)-10(x-8)>0

解答

(x^{2} - 12 x + 32) (x - 1) - 10 (x - 8) > 0

- 1 < x < 6 or x > 8

间隔符号

(- 1, 6) \cup (8, \infty)

求解步骤

(x^{2} - 12 x + 32) (x - 1) - 10 (x - 8) > 0

分解

(x^{2} - 12 x + 32) (x - 1) - 10 (x - 8) : (x - 8) (x + 1) (x - 6)

(x - 8) (x + 1) (x - 6) > 0

确定区间

- 1 < x < 6 or x > 8

\frac{6}{x - 2} > - x - 3

\frac{( - 2 x - 1 )}{( x - 2 )} + \frac{( x + 1 )}{x} \leq - \frac{1}{2}

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

- x^{2} + 2 x + 8 => 0

\frac{( x - 3 )}{( 2 x - 1 )} \geq 4