(x+2)(x+3)(x-4)(x-5)=44只能用二元一次方程求解，请写出详细过程
热心网友
(x+2)(x+3)(x-4)(x-5)=44[(x+2)(x-4)][(x+3)(x-5)]=44[x^2 - 2x -8] * [ x^2 -2x -15 ] = 44设 x^2 - 2x = y(y-8)(y-15)=44y^2 - 23y + 76 = 0(y-4)(y-19)=0x^2 - 2x - 4 =0或者x^2- 2x - 19 =0（x-1)^2 =3或者(x-1)^2 = 18x= 1±√3或者x= 1±3√2