Question

[Đỗ Văn Hoài Tuân]

Giai pt: (x-a)/(b+c) + (x-b)/(c+a) + (x-c)/(a+b) = 3x/(a+b+c)

Answer 1

[Hoàng Phúc]

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

\(\Leftrightarrow\frac{x-a}{b+c}-1+\frac{x-b}{c+a}-1+\frac{x-c}{a+b}-1=\frac{3x}{a+b+c}-3\)

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

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

\(\Leftrightarrow\frac{x-a-b-c}{b+c}+\frac{x-a-b-c}{c+a}+\frac{x-a-b-c}{a+b}-\frac{3\left(x-a-b-c\right)}{a+b+c}=0\)

\(\Leftrightarrow\left(x-a-b-c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}-\frac{3}{a+b+c}\right)=0\left(1\right)\)

Đặt \(A=\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}-\frac{3}{a+b+c}\)

Ta có: \(\left(1\right)\Leftrightarrow\orbr{\begin{cases}x-a-b-c=0\\\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}-\frac{3}{a+b+c}=0\end{cases}}\)

Xét 2 TH:

+\(A\ne0=>x-a-b-c=0=>x=a+b+c\),đây là 1 nghiệm của pt

+\(A=0\) thì phương trình có vô số nghiệm nằm trong tập hợp số thực R

Vậy........................................