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_数学解答

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

人气：477 ℃　时间：2020-01-29 21:57:51

解答

(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.