Question

計算

x-y

x+y

+

y-z

y+z

+

z-x

z+x

+

(x-y)(y-z)(z-x)

(x+y)(y+z)(z+x)

=（　　）

• A、-1
• B、0
• C、1
• D、2

Answer 1

考點：分式的加減法

專題：

分析：先通分移項得到

(x-y)(y+z)(z+x)

(x+y)(y+z)(z+x)

+

(x-y)(y-z)(z-x)

(x+y)(y+z)(z+x)

+

(y-z)(x+y)(z+x)

(x+y)(y+z)(z+x)

+

(z-x)(x+y)(y+z)

(x+y)(y+z)(z+x)

，再兩兩相加得到

2z(x-y)

(y+z)(z+x)

-

2z(x-y)

(y+z)(z+x)

，計算即可求解．

解答：解：

x-y

x+y

+

y-z

y+z

+

z-x

z+x

+

(x-y)(y-z)(z-x)

(x+y)(y+z)(z+x)

=

(x-y)(y+z)(z+x)

(x+y)(y+z)(z+x)

+

(y-z)(x+y)(z+x)

(x+y)(y+z)(z+x)

+

(z-x)(x+y)(y+z)

(x+y)(y+z)(z+x)

+

(x-y)(y-z)(z-x)

(x+y)(y+z)(z+x)

=

(x-y)(y+z)(z+x)

(x+y)(y+z)(z+x)

+

(x-y)(y-z)(z-x)

(x+y)(y+z)(z+x)

+

(y-z)(x+y)(z+x)

(x+y)(y+z)(z+x)

+

(z-x)(x+y)(y+z)

(x+y)(y+z)(z+x)

=

2z(x-y)

(y+z)(z+x)

-

2z(x-y)

(y+z)(z+x)

=0．
故選B．

點評：考查了分式的加減法，本題直接計算運算量較大，在解題過程中要根據(jù)規(guī)律靈活運用，從而簡便計算．