Question

Cho a,b,c \(\ne\)0 và \(\dfrac{1}{a}\)+\(\dfrac{1}{b}\)+\(\dfrac{1}{c}\)\(\ne\)0

Tìm x theo a,b,c

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

Answer 1

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

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

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

Mà \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ne0\) nên \(x-a-b-c=0\Rightarrow x=a+b+c\)

Vậy nghiệm của PT là \(x=a+b+c\)