求方程x(x+1)(x+2)(x+3)=24的解
问题描述：
求方程x(x+1)(x+2)(x+3)=24的解
答
[x(x+3)][(x+1)(x+2)]=24
(x²+3x)(x²+3x+2)-24=0
(x²+3x)²+2(x²+3x)-24=0
(x²+3x+6)(x²+3x-4)=0
(x²+3x+6)(x+4)(x-1)=0
x²+3x+6=0无解
x+4=0,x=-4
x-1=0,x=1
所以x=-4或x=1
答
[x(x+3)][(x+1)(x+2)]=24
(x²+3x)(x²+3x+2)-24=0
(x²+3x)²+2(x²+3x)-24=0
(x²+3x+6)(x²+3x-4)=0
(x²+3x+6)(x+4)(x-1)=0
x²+3x+6=0无解
x+4=0,x=-4
x-1=0,x=1
故x=-4或x=1
答
[x(x+3)][(x+1)(x+2)]=24
(x²+3x)[(x²+3x)+2]-24=0
(x²+3x)²+2(x²+3x)-24=0
(x²+3x+6)(x²+3x-4)=0
x²+3x+6=0,判别式小于0,无解
x²+3x-4=0
(x+4)(x-1)=0
x=-4,x=1