(((x^2-5)(x^2-4)))/((x-1))<= 0

解答

\frac{( ( x ^{2} - 5 ) ( x ^{2} - 4 ) )}{( x - 1 )} \leq 0

x \leq - 5 or - 2 \leq x < 1 or 2 \leq x \leq 5

间隔符号

(- \infty, - 5] \cup [- 2, 1) \cup [2, 5]

十进制

x \leq - 2.23606 \dots or - 2 \leq x < 1 or 2 \leq x \leq 2.23606 \dots

求解步骤

\frac{( x ^{2} - 5 ) ( x ^{2} - 4 )}{x - 1} \leq 0

分解

\frac{( x ^{2} - 5 ) ( x ^{2} - 4 )}{x - 1} : \frac{( x + 2 ) ( x - 2 ) ( x + 5 ) ( x - 5 )}{x - 1}

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

确定区间

x \leq - 5 or - 2 \leq x < 1 or 2 \leq x \leq 5

\frac{( x ^{2} - 5 x + 12 )}{( x ^{2} - 4 x + 5 )} > 3

\frac{( 2 x - x ^{2} )}{( ( 3 + x ) ( 5 - x ) )} < 0

\frac{1}{( x - 1 )} > \frac{1}{( x + 1 )}

\frac{( x - 1 ) ( x - 5 )}{( x - 3 )} > 0

\frac{1}{x} + \frac{1}{x ^{2}} > 0