已知(x+2)^2+|x+y+5|=0,求3x^2y-[2x^2y-(2xy-x^2y)-4x^2]-xy
来源:百度知道 编辑:UC知道 时间:2024/05/31 11:58:39
(x+2)^2+|x+y+5|=0,
x+2=0,x+y+5=0
x=-2,y=-3
3(x^2)y-2(x^2)y+2xy-(x^2)y+4x^2-xy
=4x^2+xy
=4*(-2)^2+(-2)*(-3)
=16+6
=22
(x+2)^2+|x+y+5|=0
x+2=0,x+y+5=0
x=-2,y=-3
3x^2y-[2x^2y-(2xy-x^2y)-4x^2]-xy
=xy+4x^2
=x(y+4x)
=-2*(-3-2*4)
=22
完全平方+绝对值=0
则x+2=0,x+y+5=0
x=-2,y=-3
3x^2y-[2x^2y-(2xy-x^2y)-4x^2]-xy
=3x^2y-2x^2y+2xy-x^2y+4x^2-xy
=xy+4x^2
=(-2)*(-3)+4*(-2)^2
=6+16=22
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