Đáp án: $x=\dfrac{a^2+b^2+c^2+\dfrac12(ab+bc+ca)}{a+b+c}$
Giải thích các bước giải:
Ta có:
$\dfrac{x-a}{bc}+\dfrac{x-b}{ca}+\dfrac{x-c}{ab}=\dfrac12(\dfrac1a+\dfrac1b+\dfrac1c)$
$\to\dfrac{a(x-a)+b(x-b)+c(x-c)}{abc}=\dfrac12(\dfrac1a+\dfrac1b+\dfrac1c)$
$\to\dfrac{ax-a^2+bx-b^2+cx-c^2}{abc}=\dfrac12(\dfrac1a+\dfrac1b+\dfrac1c)$
$\to\dfrac{x(a+b+c)-(a^2+b^2+c^2)}{abc}=\dfrac12(\dfrac1a+\dfrac1b+\dfrac1c)$
$\to\dfrac{x(a+b+c)}{abc}-\dfrac{a^2+b^2+c^2}{abc}=\dfrac12\cdot\dfrac{ab+bc+ca}{abc}$
$\to\dfrac{x(a+b+c)}{abc}=\dfrac{a^2+b^2+c^2}{abc}+\dfrac12\cdot\dfrac{ab+bc+ca}{abc}$
$\to x(a+b+c)=a^2+b^2+c^2+\dfrac12(ab+bc+ca)$
$\to x=\dfrac{a^2+b^2+c^2+\dfrac12(ab+bc+ca)}{a+b+c}$
