\(\dfrac{\left(4x-1\right)}{15}=\dfrac{\left(x+2\right)}{5}\)
\(\dfrac{\left(4x-1\right)}{15}-\dfrac{\left(x+2\right)}{5}=0\)
\(\dfrac{\left(4x-1\right)}{15}-\dfrac{3\left(x+2\right)}{3\times5}=0\)
\(\dfrac{4x-1}{15}-\dfrac{3x+6}{15}=0\)
\(4x-1-3x-6=0\)
\(x-7=0\)
\(x=7\)
\(\dfrac{4x-1}{15}=\dfrac{x+2}{5}\Rightarrow4x-1=\dfrac{15}{5}.\left(x+2\right)\)
\(\Rightarrow4x-1=3.\left(x+2\right)\)
\(\Rightarrow4x-1=3x+6\)
\(\Rightarrow x=7\)
\(\dfrac{4x-1}{15}=\dfrac{x+2}{5}\)
\(\Leftrightarrow\dfrac{4x-1}{15}=\dfrac{3\left(x+2\right)}{15}\)
\(\Leftrightarrow4x-1=3x+6\)
\(\Leftrightarrow4x-3x=6+1\)
\(\Leftrightarrow x=7\)
Vậy \(x=7\)
