《2025年全国重点高中提前招生同步强化全真试卷八年级数学上册》

19. 设$a,b,c$满足$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a + b + c}$.
求证:$\frac{1}{a^{2025} + b^{2025} + c^{2025}}=\frac{1}{a^{2025}}+\frac{1}{b^{2025}}+\frac{1}{c^{2025}}$.

答案： 19.证明:$\because \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{a + b + c}$,$\therefore abc + a^{2}c + a^{2}b + b^{2}c + ab^{2} + ac^{2} + abc + abc = 0$,即$(a + b)(b + c)(c + a) = 0$,$\therefore a + b,b + c,c + a$至少有一个为零.不妨设$a + c = 0$,$\therefore a^{2025} + b^{2025} + c^{2025} = b^{2025}$,右$= \frac{1}{b^{2025}}$,左$=$右,$\therefore a^{2025} + b^{2025} + c^{2025} = \frac{1}{a^{2025}} + \frac{1}{b^{2025}} + \frac{1}{c^{2025}}$.

20. (北京市竞赛)已知$\frac{x}{y + z + u}=\frac{y}{z + u + x}=\frac{z}{u + x + y}=\frac{u}{x + y + z}$,求$\frac{x + y}{z + u}+\frac{y + z}{u + x}+\frac{z + u}{x + y}+\frac{u + x}{y + z}$的值.

答案： 20.解:由已知得$\frac{x + y + z + u}{y + z + u} = \frac{x + y + z + u}{z + u + x} = \frac{x + y + z + u}{u + x + y} = \frac{x + y + z + u}{x + y + z}$
(1)如果分子$x + y + z + u \neq 0$,则由分母推得$x = y = z = u$.此时$\frac{x + y}{z + u} + \frac{y + z}{u + x} + \frac{u + x}{x + y} + \frac{x + y}{y + z} = 1 + 1 + 1 + 1 = 4$.
(2)如果分子$x + y + z + u = 0$,则$x + y = - (z + u),y + z = - (u + x)$.此时$\frac{x + y}{z + u} + \frac{y + z}{u + x} + \frac{u + x}{x + y} + \frac{x + y}{y + z} = - 1 - 1 - 1 - 1 = - 4$.