Question

[Trần Bảo Hân]

Cho abc (ab + bc + ca) khác 0. Giải phương trình ẩn x.

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

Answer 1

[Akai Haruma]

Lời giải:

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

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

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

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

Vì $abc(ab+bc+ac)\neq 0\Rightarrow \frac{ab+bc+ac}{abc}\neq 0$

$\Leftrightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\neq 0(2)$

Từ $(1);(2)\Rightarrow x-a-b-c=0\Rightarrow x=a+b+c$