\(\dfrac{25}{30}=\dfrac{2x+3}{6}\)
\(\Rightarrow2x+3=\dfrac{25\cdot6}{30}=5\)
\(\Rightarrow2x=2\)
\(\Rightarrow x=1\)
Ta có: \(\dfrac{25}{30}=\dfrac{2x+3}{6}\)
\(\Leftrightarrow\dfrac{2x+3}{6}=\dfrac{5}{6}\)
\(\Leftrightarrow2x+3=5\)
\(\Leftrightarrow2x=2\)
hay x=1
Vậy: x=1