ĐKXĐ: \(x\ne\dfrac{5}{2}\)
\(\dfrac{x}{2}-\dfrac{7}{5}=\dfrac{x}{2x-5}\)
=>\(\dfrac{5x-14}{10}=\dfrac{x}{2x-5}\)
=>\(\left(5x-14\right)\left(2x-5\right)=10x\)
=>\(10x^2-25x-28x+70-10x=0\)
=>\(10x^2-63x+70=0\)
\(\Delta=\left(-63\right)^2-4\cdot10\cdot70=1169>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left[{}\begin{matrix}x=\dfrac{63-\sqrt{1169}}{2\cdot10}=\dfrac{63-\sqrt{1169}}{20}\left(nhận\right)\\x=\dfrac{63+\sqrt{1169}}{20}\left(nhận\right)\end{matrix}\right.\)
\(\dfrac{x}{2}-\dfrac{7}{5}=\dfrac{x}{2x-5}\left(x\ne\dfrac{5}{2}\right)\\ \Leftrightarrow\dfrac{5x-14}{10}=\dfrac{x}{2x-5}\\ \Leftrightarrow\left(5x-14\right)\left(2x-5\right)=10x\\ \Leftrightarrow10x^2-28x-25x+70=10x\\ \Leftrightarrow10x^2-63x+70=0\\ \Leftrightarrow10\left(x^2-\dfrac{63}{10}x+7\right)=0\\ \Leftrightarrow10\left[x^2-2\cdot x\cdot\dfrac{63}{20}+\left(\dfrac{63}{20}\right)^2-\dfrac{1169}{400}\right]=0\\ \Leftrightarrow\left(x-\dfrac{63}{20}\right)^2-\dfrac{1169}{400}=0\\ \Leftrightarrow\left(x-\dfrac{63}{20}\right)^2=\dfrac{1169}{400}\\ TH1:x-\dfrac{63}{20}=\sqrt{\dfrac{1169}{400}}=\dfrac{\sqrt{1169}}{20}\\ \Leftrightarrow x=\dfrac{63+\sqrt{1169}}{20}\\ TH2:x-\dfrac{63}{20}=-\sqrt{\dfrac{1169}{400}}=\dfrac{-\sqrt{1169}}{20}\\ \Leftrightarrow x=\dfrac{63-\sqrt{1169}}{20}\)
