Đáp án: $x=a+b+c$
Giải thích các bước giải:
Ta có:
$\dfrac{x-b-c}{a}+\dfrac{x-c-a}{b}+\dfrac{x-a-b}{c}=3$
$\to (\dfrac{x-b-c}{a}-1)+(\dfrac{x-c-a}{b}-1)+(\dfrac{x-a-b}{c}-1)=0$
$\to \dfrac{x-b-c-a}{a}+d\frac{x-c-a-b}{b}+\dfrac{x-a-b-c}{c}=0$
$\to \dfrac{x-a-b-c}a+\dfrac{x-a-b-c}b+\dfrac{x-a-b-c}c=0$
$\to (x-a-b-c)(\dfrac1a+\dfrac1b+\dfrac1c)=0$
Vì $abc(ab+bc+ca)\ne 0$
$\to ab+bc+ca\ne 0, abc\ne 0$
$\to \dfrac{ab+bc+ca}{abc}\ne 0$
$\to \dfrac1a+\dfrac1b+\dfrac1c\ne 0$
$\to x-a-b-c=0$
$\to x=a+b+c$
