(1-2x²)/[2x√(1-x²)]=[√(1-x²)-x]/[√(1-x²)+x]
(1-2x²)[√(1-x²)+x]=[2x√(1-x²)][√(1-x²)-x]
[√(1-x²)+x][√(1-x²)-x][√(1-x²)+x]=[2x√(1-x²)][√(1-x²)-x]
√(1-x²)-x=0 ①
[√(1-x²)+x]²=2x√(1-x²) ②
解①得:√(1-x²)=x
1-x²=x² 2x²=1 x=±√2/2
解②得:(1-x²)+2x√(1-x²)+x²=2x√(1-x²) 得出 1=0 方程无解
经检验,x=√2/2是原方程的根 ,x=-√2/2是增根(舍去)