你這兩道不是方程,是代數(shù)式,應(yīng)該是分解因式吧?
a²-2ab+b²-3a+3b-10
=(a-b)²-3(a-b)-10
=[(a-b)+2]*[(a-b)-5]
=(a-b+2)(a-b-5)
（x+1）（x-1）-x-1
=（x+1）（x-1）-(x+1)
=(x+1)(x-1-1)
=(x+1)(x-2)第二題（x+1）（x-1）-x-10，最后不是-1是-10（x+1）（x-1）-x-10
=x²-1-x-10
=x²-x-11
=x²-x+1/4-1/4-11
=(x-1/2)²-45/4
=(x-1/2)²-[2分之3根號(hào)5）²
=(x-1/2+2分之3根號(hào)5)(x-1/2-2分之3根號(hào)5)