\(\dfrac{6x-1}{3x+2}=\dfrac{2x+5}{x-3}\) ĐKXĐ: \(x\ne\dfrac{-2}{3};x\ne3\)
\(\Rightarrow\left(6x-1\right)\left(x-3\right)=\left(2x+5\right)\left(3x+2\right)\)
\(\Leftrightarrow6x^2-18x-x+3=6x^2+4x+15x+10\)
\(\Leftrightarrow6x^2-18x-x-6x^2-4x-15x=10-3\)
\(\Leftrightarrow-38x=7\)
\(\Leftrightarrow x=\dfrac{-7}{38}\)
\(\dfrac{6x-1}{3x+2}=\dfrac{2x+5}{x-3}\left(ĐKXĐ:x\ne\dfrac{-2}{3};x\ne3\right)\)
\(\Leftrightarrow\dfrac{\left(6x-1\right)\left(x-3\right)}{\left(3x+2\right)\left(x-3\right)}=\dfrac{\left(2x+5\right)\left(3x+2\right)}{\left(3x+2\right)\left(x-3\right)}\)
\(\Rightarrow\left(6x-1\right)\left(x-3\right)=\left(2x+5\right)\left(3x+2\right)\)
\(\Leftrightarrow6x^2-18x-x+3=6x^2+4x+15x+10\)
\(\Leftrightarrow6x^2-19x+3=6x^2+19x+10\)
\(\Leftrightarrow6x^2-6x^2-19x-19x=10-3\)
\(\Leftrightarrow-38x=7\)
\(\Leftrightarrow x=\dfrac{-7}{38}\) (thỏa mãn ĐKXĐ)
Vậy \(x=\dfrac{-7}{38}\) thì hai biểu thức \(\dfrac{6x-1}{3x+2}\) và \(\dfrac{2x+5}{x-3}\) bằng nhau
