Question

已知实数x，y满足(x+√(x^2+1))(y+√(y^2+1))=1,求x+y的值

已知实数x，y满足(x+√(x^2+1))(y+√(y^2+1))=1,求x+y的值过程尽量详细些。谢谢

Answer 1

x+√(x^2+1)=1/[√(x^2+1)-x]
所以(x+√(x^2+1))(y+√(y^2+1))=1等价唤睁于
(y+√(y^2+1))/[√(x^2+1)-x]=1
即[√(x^2+1)-x]=y+√(y^2+1)
所以x+y=√(x^2+1)+√(y^2+1)
两边平方得到（x^2+2xy+y^2)=x^2+1+y^2+1+2√[(x^2+1)(y^2)]
整理唯启得到1-xy=√和山岁[(x^2+1)(y^2)]
再平方得打1+x^2y^2-2xy=x^2+y^2+1+x^2y^2
整理得到x^2+y^2+2xy=0
即(x+y)^2=0
所以x+y=0