Question

求解一道恒等变形数学题已知实数x,y 满足[(1+x^2)^(1/2)-x][(1+y^2)^(1/2)-y]=1试猜想

Answer 1

x+y=0!

两边分别乘[(1+x^2)^(1/2)+x]与[(1+y^2)^(1/2)+y]可得两式：

(1+y^2)^(1/2)-y=(1+x^2)^(1/2)+x

(1+x^2)^(1/2)-x=(1+y^2)^(1/2)+y

两式相加：-（x+y）=x+y

所以：x+y=0