lim （x,y）->(0,0) xy/[根号下(xy+1)]-1的值为

因为根号下（x+2）+(y-根号下3）平方=0
根号的得数大于等于0 平方的得数大于等于0
所以x+2=0 y-根号下3=0
所以X=-2 Y=根号3
XY=-2*根号3
XY=-2根号3

xy/[√(xy+1)-1]
=xy[√(xy+1)-1]/(xy+1-1)
=[√(xy+1)-1]
=1-1
=0

题目应该是：x²-4x+y²+6y+√(z+1)+13=0
∴x²-4x+y²+6y+√(z+1)+4+9=0
∴（x²-4x+4）+（y²+6y+9）+√(z+1）=0
∴（x-2）²+（y+3）²+√(z+1)=0
∴x=2,y=-3,z=-1
∴（xy）²=(-6)²=36

因为根号大于等于0
两个大于等于0的数相加等于0
说明每个数都等于0
所以 x + 2 = 0 ，y - √3 = 0
所以 x = -2 ，y = √3
所以 xy = -2√3

x^2+(y^2)/2=1,
x^2+[(1/√2)y]^2=1,
设x=cosA,y=√2sinA,
因x＞0,y＞0,不妨设0＜A＜π/2,
x√(1+y^2)=cosA√[1+2(sinA)^2]
=√{(cosA)^2[1+2(sinA)^2]}
=√{[1-(sinA)^2][1+2(sinA)^2]}
=√[1+(sinA)^2-2(sinA)^4]
=√{1+(1/√2)[√2(sinA)^2]-[√2(sinA)^2]^2}
=√{-[√2(sinA)^2-1/(2√2)]^2+[1/(2√2)]^2+1}
=√{-[√2(sinA)^2-√2/4]^2+9/8}
=√{-2[(sinA)^2-1/4]^2+9/8}
=√{-2[(sinA+1/2)(sinA-1/2)]^2+9/8}
≤√(9/8)=(3/4)√2
当sinA=1/2,即x=y=√2/2时,原式取最大值(3/4)√2.

xy/[√(xy+1)-1]
=xy[√(xy+1)-1]/(xy+1-1)
=[√(xy+1)-1]
=1-1
=0

xy/[√(xy+1)-1]
=xy[√(xy+1)-1]/(xy+1-1)
=[√(xy+1)-1]
=1-1
=0