Question

Giải phương trình:

\(\dfrac{x-a}{b+c}+\dfrac{x-b}{c+a}+\dfrac{x-c}{a+b}=\dfrac{3x}{a+b+c}\)(a,b,c là tham số)

Answer 1

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

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

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

Nếu \(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}-\dfrac{3}{a+b+c}=0\) thì PT có nghiệm với mọi \(x\in R\)

Nếu \(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}-\dfrac{3}{a+b+c}\ne0\) thì PT có nghiệm là \(x=a+b+c\)