Question

(X+2)(X+3)(X-4)(X-5)=144

Answer 1

(X+2)(X+3)(X-4)(X-5)=144

左边分组：[(X+2)(X-4)][(X+3)(X-5)]=144

整理得：（X^2-2X-8)(X^2-2X-15)=144

换元：设X^2-2X-8=Y

则原方程可化为：Y（Y-7)=144

解得Y=16或Y=-9

再解X^2-2X-8=16和X^2-2X-8=-9

不再赘述.