已知:| x + y + 1| +| xy - 3 | = 0,求代数式xy3 + x3y 的值.

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∵| x + y + 1|≥0,| xy - 3 |≥0
| x + y + 1| +| xy - 3 | = 0,
∴x+y+1=0,即x+y=-1
xy=3xy3 + x3y
=xy(x²+y²)
=yx[(x+y)²-2xy]
=3×(1-6)
=-15======以下答案可供参考======
供参考答案1:
| x + y + 1| +| xy - 3 | = 0,
x+y+1=0
xy-3=0
x+y=-1
xy=3xy3 + x3y
=xy(y^2+x^2)
=xy[(x+y)^2-2xy)
=3[(-1)^2-2*3]
=2*(-5)
=-10供参考答案2:
绝对值相加为零.必然两者为零.解出XY 代入即可
供参考答案3:
x+y+1=0
------(x+y)^2=1------x^2+y^2+2xy=1xy-3=0
------xy=3所以--------x^2+y^2=1-6=-5
xy^3+x^3y=xy(x^2+y^2)=3*(-5)=-15