已知a,b,c是正数，解关于x的方程

${\displaystyle \frac{x-a-b}{c}}+{\displaystyle \frac{x-b-c}{a}}+{\displaystyle \frac{x-c-a}{b}}$ = $3$

解法一：原方程化为

x$({\displaystyle \frac{1}{a}}+{\displaystyle \frac{1}{b}}+{\displaystyle \frac{1}{c}})$ = ${\displaystyle \frac{b+c}{a}}+{\displaystyle \frac{c+a}{b}}+{\displaystyle \frac{a+b}{c}}+3$ $$①

因为 ${\displaystyle \frac{b+c}{a}}+{\displaystyle \frac{c+a}{b}}+{\displaystyle \frac{a+b}{c}}+3$

=$({\displaystyle \frac{b+c}{a}}+1)+({\displaystyle \frac{c+a}{b}}+1)+({\displaystyle \frac{a+b}{c}}+1)$

=${\displaystyle \frac{b+c+a}{a}}+{\displaystyle \frac{c+a+b}{b}}+{\displaystyle \frac{a+b+c}{c}}$

所以方程①化为$$x$({\displaystyle \frac{1}{a}}+{\displaystyle \frac{1}{b}}+{\displaystyle \frac{1}{c}})=(a+b+c)({\displaystyle \frac{1}{a}}+{\displaystyle \frac{1}{b}}+{\displaystyle \frac{1}{c}})$$$②

因为a,b,c都是正数，所以${\displaystyle \frac{1}{a}}+{\displaystyle \frac{1}{b}}+{\displaystyle \frac{1}{c}}>0$,从而在方程②两边同时除以${\displaystyle \frac{1}{a}}+{\displaystyle \frac{1}{b}}+{\displaystyle \frac{1}{c}}$，得x = a+b+c。

解法二：原方程化为

$({\displaystyle \frac{x-a-b}{c}}-1)+({\displaystyle \frac{x-b-c}{a}}-1)+({\displaystyle \frac{x-c-a}{b}}-1)=0$

得${\displaystyle \frac{x-a-b-c}{c}}+{\displaystyle \frac{x-b-c-a}{a}}+{\displaystyle \frac{x-c-a-b}{b}}=0$

提取公因式x-a-b-c,得

$$$$$$$$$$$$$$$(x-a-b-c)({\displaystyle \frac{1}{c}}+{\displaystyle \frac{1}{a}}+{\displaystyle \frac{1}{b}})=0$

因为${\displaystyle \frac{1}{a}}+{\displaystyle \frac{1}{b}}+{\displaystyle \frac{1}{c}}>0$,所以$x-a-b-c=0,即$x=a+b+c。 故方程的解为$x=a+b+c$。

解法三：假设${\displaystyle \frac{x-a-b}{c}}=1,{\displaystyle \frac{x-b-c}{a}}=1,{\displaystyle \frac{x-c-a}{b}}=1,$则每个分式都可以得到$x=a+b+c。$ $$又题中方程是一次方程，它只有一个解，故方程的解是$x=a+b+c$。