Question

先化简,再求代数式(x-3)(x+2)+(x+3)^2-2(x-y)(x+y)的值,其中x+2的绝对值+(y-1)^2=0

先化简,再求代数式(x-3)(x+2)+(x+3)^2-2(x-y)(x+y)的值,其中x+2的绝对值+(y-1)^2=0肿么写啊啊啊

Answer 1

(x-3)(x+2)+(x+3)^2-2(x-y)(x+y)

=x^2-x-6+x^2+6x+9-2x^2+2y^2
=5x+2y^2+3
由 |x+2|+(y-1)^2=0

x=-2，y=1
(x-3)(x+2)+(x+3)^2-2(x-y)(x+y)=5x+2y^2+3=-10+2+3=-5。

Answer 2

代数式(x-3)(x+2)+(x+3)^2-2(x-y)(x+y)化简得5x+2y^2+3,
又|x+2|+(y-1)^2=0，所以x=-2,y=1,代入上式得值-5，即
(x-3)(x+2)+(x+3)^2-2(x-y)(x+y)=-5

Answer 3

|x+2|+(y-1)^2=0 => x+2=0,y-1=0
(x-3)(x+2)+(x+3)^2-2(x-y)(x+y) =(x-3)*0+ (x+3)^2-2(x-y)(x+y) = 1-2(-2-1)(-2+1)=-5

Answer 4

原式=(X^2-X-6)+(X^2+6X+9)-2(X^2-y^2)
=2y^2-5X+3
因为X+2的绝对值+（y-1）^2=0
所以X=-2,y=1
所以，原式=15