Question

1.已知x+y=-4,xy=3,求5（2分之1xy+2y）+[6x-(2xy+y-3x)]的值

2.已知（x+3）^2+|x-y+10|=0
求代数式5x^2y-[2x^2y-(3xy-xy^2)-3x^2]-2xy^2-y^2的值

Answer 1

1.已知x+y=-4,xy=3,求5(xy/2+2y)+[6x-(2xy+y-3x)]的值
5(xy/2+2y)+[6x-(2xy+y-3x)]
=5xy/2+10y+6x-2xy-y+3x
=xy/2+9y+9x
=xy/2+9(x+y)
=3/2-4=-5/2

2.已知(x+3) ²+|x-y+10|=0
求代数式5x²y-[2x²y-(3xy-xy²)-3x²]-2xy²-y²的值
(x+3)^2+|x-y+10|=0
(x+3)^2=0 |x-y+10|=0
x+3=0 x-y+10=0
x=-3 y=7
5x²y-[2x²y-(3xy-xy²)-3x²]-2xy²-y²
=5x²y-2x²y+(3xy-xy²)+3x²-2xy²-y²
=5x²y-2x²y+3xy-xy²+3x²-2xy²-y²
=3x²y+3xy-3xy²+3x²-y²
=3xy(x+1-y)+3x²-y²
=-3*3*7*(-3+1-7)+3*9-49
=545