Question

Cho a, b, c \(\ne\) 0 và \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ne0\). Giải phương trình ẩn x sau:

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

Answer 1

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

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

\(\) vì \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ne0\Rightarrow x-a-b-c=0\)

\(\Rightarrow x=a+b+c\)