一高中数学题..求解实数x，y满足x平方+y平方=1，则2xy/(x+y-1)的最大值

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均值不等式======以下答案可供参考======
供参考答案1:
x^2+y^2=1
(x+y)^2-2xy=1
2xy=(x+y)^2-1=(x+y+1)(x+y-1)
2xy/(x+y-1)
={(x+y)^2-1}/(x+y-1)
={(x+y+1)(x+y-1)}/(x+y-1)
=x+y+1
x^2+y^2=1
x=sina y-cosa
x+y=sina+cosa=√2sin(a+∏/4)
MAX(x+y)=√2
2xy/(x+y-1)
=x+y+1的最大值√2+1供参考答案2:
因为x²+y²=1,
所以2xy=(x+y)²-(x²+y²)=(x+y)²-1
2xy/(x+y-1)
=[(x+y)²-1]/(x+y-1)=[(x+y-1)(x+y+1)]/(x+y-1)=x+y+1所以x²=y²=1/2时(0＜x,y)
上式有最大值为“（根号2）+1”
供参考答案3:
2xy/(x+y-1)
=[x^2+y^2+2xy-1]/[x+y-1]=[(x+y)^2-1]/[x+y-1]=[(x+y+1)*(x+y-1)/[x+y-1]=x+y+1=√[x+y]^2 +1=√[x^2+y^2+2xy]+1≤√[x^2+y^2+x^2+y^2]+1=√2+1