MSC=24
Ta có: \(\frac{x+5}{4}-\frac{2x-3}{3}=\frac{6x-1}{8}+\frac{2x-1}{12}\)
⇔\(\frac{x+5}{4}-\frac{2x-3}{3}-\frac{6x-1}{8}-\frac{2x-1}{12}=0\)
⇔\(\frac{6\left(x+5\right)}{4}-\frac{8\left(2x-3\right)}{24}-\frac{3\left(6x-1\right)}{24}-\frac{2\left(2x-1\right)}{24}=0\)
⇔\(6x+30-\left(16x-24\right)-\left(18x-3\right)-\left(4x-2\right)=0\)
\(\Leftrightarrow6x+30-16x+24-18x+3-4x+2=0\)
\(\Leftrightarrow-32x+59=0\)
\(\Leftrightarrow-32x=-59\)
hay \(x=\frac{-59}{-32}=\frac{59}{32}\)
Vậy: \(x=\frac{59}{32}\)
