解不等式(x+1)(x+2)(x+3)(x+4)<120

解答；
不妨组合(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
令y=x^2+5x，简化得............................................(a)
(y+4)(y+6)<120
y^2+10y-96<0
(y-6)(y+16)<0
有 -16代入(a)，得
-16解得 -3

（x-1)(x+1)(x+2)(x+3)<120
(x+1)(x+2)(x+3)>120/(1-x)
题目化简
120/(x+4)>120/(1-x)
x+4<1-x
x<-3/2

(x^2+5x+4)(x^2+5x+6)<120 (x^2+5x)^2+10(x^2+5x)-96<0 (x^2+5x+16)(x^2+5x-6)<0 (x^2+5x+16)(x+6)(x-1)<0 -6