Question

已知x²+y²-4x+y+4又4分之1=0，求xy的值

Answer 1

解：
x²+y²-4x+y+4又4分之1
= x²-4x+4+y²+y+1/4
= (x-2)²+(y+1/2)²
= 0
所以 x-2 =0，y+1/2 =0
得 x =2， y = -1/2
xy = 2×(-1/2) = -1

　
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Answer 2

解：∵x²+y²-4x+y+4又4分之1=0
∴ x²－4x＋4＋y²＋y＋1/4＝0
﹙x－2﹚²＋﹙y＋1/2﹚²＝0
∵ ﹙x－2﹚²≥0, ﹙y＋1/2﹚²≥0
∴ x－2＝0, x＝2
y＋1/2＝0, y＝﹣1/2
∴ x·y＝2×﹙﹣1/2﹚＝﹣1.

Answer 3

x²+y²-4x+y+4+1/4=0
(x²-4x+4)+(y²+y+1/4)=0
(x-2)²+(y+1/2)²=0
x-2=0;
y+1/2=0
∴x=2；y=-1/2

Answer 4

x^2+y^2-4x+y+4+1/4
=(x^2-4x+4)+(y^2+y+1/4)
=(x-2)^2+(y+1/2)^2=0
所以x=2 y=-1/2
所以xy=-1

Answer 5

x²+y²-4x+y+4又4分之1=0
(x²-4x+4)+(y²+y+4分之1)=0
(x-2)²+(y+2分之1)²=0
x-2=0，y+2分之1=0
x=2，y=-2分之1
xy=-1

Answer 6

x²+y²-4x+y+4又4分之1
= x²-4x+4+y²+y+1/4
= (x-2)²+(y+1/2)²
= 0