Phương trình có mỗi một vế sao giải được bạn
Ta có :\(\frac{x-b-c}{a}+\frac{x-c-a}{b}+\frac{x-a-b}{c}=3\)
\(\Rightarrow\frac{x-b-c}{a}+\frac{x-c-a}{b}+\frac{x-a-b}{c}-3=0\)
\(\Rightarrow\left(\frac{x-b-c}{a}-1\right)+\left(\frac{x-c-a}{b}-1\right)+\left(\frac{x-a-b}{c}-1\right)=0\)
\(\Leftrightarrow\frac{x-a-b-c}{a}+\frac{x-a-b-c}{b}+\frac{x-a-b-c}{c}=0\)
\(\Leftrightarrow\left(x-a-b-c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=0\) (1)
mà \(ab+bc+ca\ne0\)
\(\Rightarrow\frac{ab}{abc}+\frac{bc}{abc}+\frac{ca}{abc}\ne0\) hay \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ne0\) (2)
Từ (1)(2) => x-a-b-c=0
=> x=a+b+c
Vậy ....
