丨x-3丨+(x-y+1)^2=0
則x=3,y=4
所以x^2y+xy^2+y^3/4
=3^2*4+3*4^2+4^3/4
=36+48+16
=84+16
=100
若丨x-3丨+(x-y+1)^2=0,計算根號X^2Y + XY^2 +4分之y^3=
若丨x-3丨+(x-y+1)^2=0,計算根號X^2Y + XY^2 +4分之y^3=
根號X2Y + XY2 +Y2
根號X2Y + XY2 +Y2
數(shù)學(xué)人氣:528 ℃時間:2020-05-11 16:42:45
優(yōu)質(zhì)解答
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