\(2+\dfrac{x}{5}=\dfrac{2x-1}{5}-\dfrac{3x+2}{4}\)
\(\Leftrightarrow\dfrac{40}{20}+\dfrac{4x}{20}=\dfrac{4\left(2x-1\right)}{20}-\dfrac{5\left(3x+2\right)}{20}\)
\(\Rightarrow40+4x=8x-4-15x-10\)
\(\Leftrightarrow4x-8x+15x=-4-10-40\)
\(\Leftrightarrow11x=-54\)
\(\Leftrightarrow x=\dfrac{-54}{11}\)
2 + \(\frac{x}{5}=\frac{2x-1}{5}-\frac{3x+2}{4}\)
\(\Leftrightarrow\frac{40}{20}+\frac{4x}{20}=\frac{4\left(2x-1\right)}{20}-\frac{5\left(3x-2\right)}{20}\)
\(\Leftrightarrow40+4x=8x-4-15x+10\)
\(\Leftrightarrow4x-8x+15x=10-4-40\)
\(\Leftrightarrow3x=-34\)
\(\Leftrightarrow x=\frac{-34}{3}\)
Vậy S = \(\left\{\frac{-34}{3}\right\}\)
\(2+\frac{x}{5}=\frac{2x-1}{5}-\frac{3x+2}{4}\)
\(\Leftrightarrow\frac{40}{20}+\frac{4x}{20}=\frac{4\left(2x-1\right)}{20}-\frac{5\left(3x+2\right)}{20}\)
\(\Leftrightarrow40+4x=8x-4-15x-10\)
\(\Leftrightarrow4x-8x+15x=-10-4-40\)
\(\Leftrightarrow9x=-54\)
\(\Leftrightarrow x=-6\)
Vậy S = \(\left\{-6\right\}\)
