x(x+1)(x+2)(x+3)=3024
[x(x+3)][(x+1)(x+2)]=3024
(x^2+3x)(x^2+3x+2)=3024
设 x^2+3x =y
y(y+2)=3024
y^2+2y-3024=0
(y+56)(y-54)=0
y1= -56
y2= 54
x^2+3x=-56
x^2+3x+56=0
(x+ 3/2)^2 -9/4 +56=0
(x+ 3/2)^2 + 53.75 =0
上面等式左端的最小值是 53.75,永远不会为0.因此无解.
x^2+3x=54
x^2+3x-54=0
(x+9)(x-6)=0
x1 = -9
x2 = 6