Question

解方程(x+3)(x-4)-(x-1)(x-2)＝20？

Answer 1

解由(x+3)(x-4)-(x-1)(x-2)＝20
得(x^2-x-12)-(x^2-3x+2)=20
即2x-14=20
即2x=34
解得x=17

Answer 2

（x+2)(x+3)(x-4)(x-5)=144
[(x+2)(x-4)][(x+3)(x-5)]=144
(x^2-2x-8)(x^2-2x-15)=144
(x^2-2x)^2-23(x^2-2x)+120=144
(x^2-2x)^2-23(x^2-2x)-24=0
[(x^2-2x)-24][(x^2-2x)+1]=0
(x-6)(x+4)(x-1)^2=0
x1=6,x2=-4,x3=1
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令t=x-1
则x+2=t+3，x+3=t+4，x-4=t-3，x-5=t-4
于是原方程变为：(t+4)(t+3)(t-3)(t-4)=144
即(t^2-9)(t^2-16)=144
即t^4-25t^2=0，亦即t^2(t+5)(t-5)=0
所以t=0，或t=-5，或t=5
则x=t+1=1，或-4，或6