Question

[Nguyen Hong Ngoc]

 

Answer 1

[Toru]

ĐKXĐ: \(x\ne2\)

\(\dfrac{x^2}{2-x}+\dfrac{3x-1}{3}=\dfrac{5}{3}\)

\(\Leftrightarrow\dfrac{x^2}{2-x}=\dfrac{5}{3}-\dfrac{3x-1}{3}\)

\(\Leftrightarrow\dfrac{x^2}{2-x}=\dfrac{6-3x}{3}\)

\(\Leftrightarrow\dfrac{x^2}{2-x}=2-x\)

\(\Rightarrow x^2=\left(2-x\right)^2\)

\(\Leftrightarrow x^2=x^2-4x+4\)

\(\Leftrightarrow4x=4\)

\(\Leftrightarrow x=1\) (tm ĐKXĐ)