2024-06-10

已知x√(1-y²)+y√(1-x²)=1,求证：x²+y²=1.

证明：方法一：

移项得：x√(1-y²)=1-y√(1-x²) 两边平方得：x²(1-y²)=1+y²(1-x²)-2y√(1-x²).x²-x²y²=1+y²-x²y²-2y√(1-x²).

1-x²-2y√(1-x²)+y²=〔√(1-x²)-y〕²=0

√(1-x²)=y.

1-x²=y²,∴x²+y²=1

方法二：设原方程为①式；令x√(1-y²)-y√(1-x²)=k    ②。

①×②得:x²(1-y²)-y²(1-x²)=k. x²-y²=k;①+②得：2x√(1-y²)=1+k=1+x²-y².即：1-y²+x²-2x√(1-y²)=0. 〔√(1-y²)-x〕²=0.

√(1-y²)=x,1-y²=x², x²+y²=1.

两种解题方法关键点：方法一：

1-x²-2y√(1-x²)+y²=√(1-x²)²-2y√(1-x²)+y²

=〔√(1-x²)-y〕²=0,

√(1-x²)-y=0,√(1-x²)=y,1-x²=y² ,x²+y²=1.

方法二：

1-y²+x²-2x√(1/y²)=〔√(1-y²)²-2x√(1-y²)+x²〕=〔√(1-y²)-x〕².

〔√(1-y²)-x〕²=0.

√(1-y²)-x=0.

√(1-y²)=x,1-y²=x²

x²+y²=1.