(2x-3)(x-4)-(x+2)(x-3)=x+8
解:x^2+10=11x
x^2-11x+10=0
(x-10)x+18=x+8
x=1
x=10
式子展开:
2x^2-8x-3x+12-(x^2-3x+2x-6)=x+8
2x^2-11x+12-x^2+x+6=x+8
x^2-10x+18=x+8
x^2-11x+10=0
(x-1)(x-10)=0
X1=1,X2=10
2x²-11x+12-(x²-x-6)=x+8
2x²-11x+12-x²+x+6=x+8
x²-11x+10=0
(x-1)(x-10)=0
x-1=0,x1=1
x-10=0,x2=10
2x²-11x+12-x²+x+6=x+8
x²-10x+18=x+8
x²-11x+10=0
(x-1)(x-10)=0
x1=1
x2=10