解不等式(X---1)(X--2)(X--3)(X--4)>120

(x-1)(x-2)(x-3)(x-4)>120
(x-1)(x-4)(x-2)(x-3)>120
(x^2-5x+4)(x^2-5x+6)>120
(x^2-5x)^2+10*(x^2-5x)+24>120
(x^2-5x)^2+10*(x^2-5x)-96>0
(x^2-5x+16)(x^2-5x-6)>0
x^2-5x+16恒大于0
所以x^2-5x-6>0
(x-6)(x+1)>0
x<-1或x>6

(X---1)(X--2)(X--3)(X--4)>120
=>(X-1)(X-2)(X-3)(X-4)>2*3*4*5
显然当x>6时成立
=>(1-x)(2-x)(3-x)(4-x)>2*3*4*5
显然当x<-1时成立

当-1<x<6时
120>0
=>(X-1)(X-2)(X-3)(X-4)>0
=>2<X<3

(X---1)(X--2)(X--3)(X--4)>120
=>(X-1)(X-2)(3-X)(4-X)>120

(X-1)(4-X)=<(1.5)^2
(X-2)(3-X)=<(1.5)^2
=>(X-1)(X-2)(3-X)(4-X)<120
故x>6 x<-1

(X-1)(X-2)(X-3)(X-4)>120
(X-1)(X-2)(X-3)(X-4)-120>0
全部展开再分解太麻烦了。
为了方便，记A=(X-1)(X-2)(X-3)(X-4)-120
则不等式为A>0
注意-1-4=-2-3，所以
A=(X-1)(X-2)(X-3)(X-4)-120
=[(X-1)(X-4)][(X-2)(X-3)]-120
=[X^2-5X+4][X^2-5X+6]