Question

二次根式数学若x^2-y^2=1,且x＞0,y＞0,求x√(1+y^2)-y√(x^2-1)的值

Answer 1

x²-y²=1

1+y²=x²

x²-1=y²

所以原式=x√x²-y√y²

=x|x|-y|y|

x>0.y>0

所以原式=x²-y²=1