Question

已知x^2+2x=2,求代数式(x-1)2+(x+3)(x-3)+(x-3)(x-1)的值

计算(a-b)^5(b-a)^3

Answer 1

(x-1)^2+(x+3)(x-3)+(x-3)(x-1)
=x^2-2x+1+x^2-9+x^2-4x+3
=3x^2-6x-5
=3(x^2-2x)-5
=3*2-5
=1
注：x^2+2x=2应为x^2-2x=2

(a-b)^5(b-a)^3
=-(a-b)^5(a-b)^3
=-(a-b)^8

Answer 2

(x-1)2+(x+3)(x-3)+(x-3)(x-1)
=x²-2x+1+x²-9+x²-4x+3
=3x²-6x+5
=3(x²-2x)+5
而条件是x^2+2x=2，没有x²-2x，是题目有错吧。