(((x-7-x^2)(x^2+6)))/((x^2-x-1))<= 0

解答

\frac{( ( x - 7 - x ^{2} ) ( x ^{2} + 6 ) )}{( x ^{2} - x - 1 )} \leq 0

x < \frac{1 - 5}{2} or x > \frac{1 + 5}{2}

间隔符号

(- \infty, \frac{1 - 5}{2}) \cup (\frac{1 + 5}{2}, \infty)

十进制

x < - 0.61803 \dots or x > 1.61803 \dots

求解步骤

\frac{( x - 7 - x ^{2} ) ( x ^{2} + 6 )}{x ^{2} - x - 1} \leq 0

因式分解

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

\frac{( - ( - x + 7 + x ^{2} ) ) ( x ^{2} + 6 )}{( x - \frac{1 + 5}{2} ) ( x - \frac{1 - 5}{2} )} \leq 0

确定区间

x < \frac{1 - 5}{2} or x > \frac{1 + 5}{2}

x (x^{2} - x) (x^{3} - x) (x^{4} - x) < 0

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

x^{3} + x^{3} - x - 1 > 0

(x^{2} - 4) (x^{2} - 9) \geq 0

m^{2} =< \frac{2}{3}