若xy-4x+y^2+16y+13=0

来源：百度知道 编辑：UC知道 时间：2024/05/30 18:54:08
过程要仔细？求x=? y=？

xy-4x+y^2+16y+13=0
(y^2+xy+1/4x^2)-1/4x^2-4x+16y+13=0
(y+1/2x)^2-(1/4x^2+4x+16)+16y+13+16=0
(y+1/2x)^2-(1/2x+4)^2+16y+29=0
(y+x+4)(y-4)+(16y+29)=0
要使方程此时有解需同时满足下面两个条件：
(y+x+4)(y-4)=0
16y+29=0

解得x=35/16，y=-29/16
即x=35/16，y=-29/16 为原方程的解。

我代入方程看过了，是这个方程的解。

x^2-4x+y^2+6y+13=0

(x^2-4x+4)+(y^2+6y+9)=0

(x-2)^2+(y+3)^2=0

x-2=0
y+3=0

x=2
y=-3

xy-4x+y^2+16y+13=0
(y^2+xy+1/4x^2)-1/4x^2-4x+16y+13=0
(y+1/2x)^2-(1/4x^2+4x+16)+16y+13+16=0
(y+1/2x)^2-(1/2x+4)^2+16y+29=0
(y+x+4)(y-4)+(16y+29)=0

(y+x+4)(y-4)=0
16y+29=0

解得x=-8，y=4或x=35/16，y=-29/16