Question

设x，y，z，w为四个互不相等的实数，并且x+

1

y

=y+

1

z

=z+

1

ω

=w+

1

x

求证：x²y²z²w²=1

Answer 1

证明：∵x+

1

y

=y+

1

z

=z+

1

ω

=w+

1

x

∴

x+ 1 y =y+ 1 z y+ 1 z =z+ 1 ω z+ 1 ω =ω+ 1 x ω+ 1 x =x+ 1 y

?

x-y= 1 z - 1 y y-z= 1 ω - 1 z z-ω= 1 x + 1 ω ω-x= 1 y - 1 y

?

(x-y)zy=y-z          ① (y-z)ωz=z-ω        ② (z-ω)xω=ω-x       ③ (ω-x)yx=x-y         ④

由①×②×③×④得，x²y²z²w²（x-y）（y-z）（z-w）（w-x）=（x-y）（y-z）（z-w）（w-x）
∵x，y，z，w互不相等
∴x²y²z²w²=1．