若x²-4x+y²+2y+5=0,求x,y的值

若x²-4x+y²+2y+5=0,求x,y的值
若x²-4x+y²+2y+5=0,求x,y的值

若x²-4x+y²+2y+5=0,求x,y的值
(x²-4x+4)+(y²+2y+1)=0
(x-2)²+(y+1)²=0
所以x-2=0,y+1=0
x=2,y=-1

x²-4x+y²+2y+5=0
x²－4x＋4＋y²＋2y＋1=0
﹙x－2﹚²＋﹙y＋1﹚²=0
x-2=0,y+1=0
x=2,y=-1

x²-4x+y²+2y+5=0
(x-2)²+(y+1)²=0
(x-2)²>=0,(y+1)²>=0;
(x-2)²=0,(y+1)²=0;
x=2,y=-1;

(x-2)²+(y+1)²=0
x=2,y=-1

x^2-4x+4+y^2+2y+1=0,所以 (x-2)62+(y+1)^2=0的x=2,y=-1

x²-4x+y²+2y+5=0
[(x-2)²-4]+[(y+1)²-1]+5=0
(x-2)²+(y+1)²=0
必有x-2=0,y+1=0
即 x=2,y=-1

x²-4x+y²+2y+5=0可以化成
（x-2）^2+(y+1)^2=0
由此可知
（x-2）^2=0
(y+1)^2=0
所以x=2
y=-1