Question

Cho \(ab+bc+ca\) giải phương trình ẩn x:

\(\frac{x-b-c}{a}+\frac{x-c-a}{b}+\frac{x-b-a}{c}=3\)

Answer 1

Ta có \(\frac{x-b-c}{a}+\frac{x-c-a}{b}+\frac{x-b-a}{c}=3\)

\(\Rightarrow\frac{x-b-c}{a}+\frac{x-c-a}{b}+\frac{x-b-a}{c}-3=0\)

\(\Leftrightarrow\left(\frac{x-b-c}{a}-1\right)+\left(\frac{x-c-a}{b}-1\right)+\left(\frac{x-b-a}{c}-1\right)=0\)

\(\Leftrightarrow\frac{x-a-b-c}{a}+\frac{x-a-b-c}{b}+\frac{x-a-b-c}{c}=0\)

\(\Leftrightarrow\left(x-a-b-c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=0\)

Vì \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ne0\) nên chỉ có

x-a-b-c=0 =>x=a+b+c

Vậy x=a+b+c