Question

[x+√(x^2+1)][y+√(y^2+1)]=1 求x+y

Answer 1

x+√(x²+1)=1/[y+√(y²+1)]
x+√(x²+1)=√(y²+1)-y

x+y=√(y²+1)-√(x²+1)
两边平方
x²+2xy+y²=y²+1+x²+1-2√(x²y²+x²+y²+1)
xy-1=-√(x²y²+x²+y²+1)

平方
x²y²-2xy+1=x²y²+x²+y²+1
x²+y²+2xy=0
(x+y)²=0
x+y=0

Answer 2

[x+√(x^2+1)][y+√(y^2+1)]=1
同乘[√(x^2+1)-x][x+√(x^2+1)][y+√(y^2+1)]=[√(x^2+1)-x]
即y+√(y^2+1)=√(x^2+1)-x
同理乘以√(y^2+1)-y
x+√(x^2+1)=√(y^2+1)-y
x+y=0