已知x2-4x-y2-10+29=0,求:x눀y눀+2x대y눀+x^4y눀的值

2024-12-20 17:48:48
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回答1:

应该是x^2-4x+y^2-10y+29=0吧
x^2-4x+y^2-10y+29=0
(x^2-4x+4)+(y^2-10y+25)=0
(x-2)^2+(y-5)^2=0
x=2,y=5
代入得
x^2y^2+2x^3y^2+x^4y^2
=x^2y^2(1+2x+x^2)
=x^2y^2(1+x)^2
=4*25*9
=900
祝学习进步!