Ta có: \(\dfrac{3x+5}{16}+\dfrac{2x+7}{8}=\dfrac{4+3x}{32}+2x\)
\(\Leftrightarrow\dfrac{2\left(3x+5\right)}{32}+\dfrac{4\left(2x+7\right)}{32}=\dfrac{4+3x}{32}+\dfrac{64x}{32}\)
Suy ra: \(6x+10+8x+28=4+3x+64x\)
\(\Leftrightarrow67x+4-14x-38=0\)
\(\Leftrightarrow53x=34\)
hay \(x=\dfrac{34}{53}\)