\(\dfrac{x-2}{4}-\dfrac{x+3}{3}=\dfrac{2x-5}{6}\\ \Leftrightarrow\dfrac{3\left(x-2\right)-4\left(x+3\right)}{12}=\dfrac{2\left(2x-5\right)}{12}\\ \Leftrightarrow3x-6-4x-12=4x-10\\ \Leftrightarrow3x-4x-4x=-10+12+6\\ \Leftrightarrow-5x=8\\ \Leftrightarrow x=\dfrac{-8}{5}\)
Vậy \(x=\dfrac{-8}{5}\) là nghiệm của pt
Ta có: \(\dfrac{x-2}{4}+\dfrac{x+3}{3}=\dfrac{2x-5}{6}\)
\(\Leftrightarrow\dfrac{3\left(x-2\right)}{12}+\dfrac{4\left(x+3\right)}{12}=\dfrac{2\left(2x-5\right)}{12}\)
\(\Leftrightarrow3x-6+4x+12=4x-10\)
\(\Leftrightarrow7x+6-4x+10=0\)
\(\Leftrightarrow3x+16=0\)
\(\Leftrightarrow3x=-16\)
\(\Leftrightarrow x=-\dfrac{16}{3}\)
Vậy: \(S=\left\{-\dfrac{16}{3}\right\}\)
