\(1+\dfrac{2x+1}{3}>\dfrac{2x-1}{6}\)
\(\Leftrightarrow6+4x+2>2x-1\)
\(\Leftrightarrow4x-2x>-1-6-2\)
\(\Leftrightarrow x>-\dfrac{9}{2}\)
Vậy S = { x/ x > \(-\dfrac{9}{2}\)}
\(1+\dfrac{2x+1}{3}>\dfrac{2x-1}{6}\)
⇔ \(\dfrac{6+2\left(2x+1\right)}{6}>\dfrac{2x-1}{6}\)
⇔ 6 + 4x + 2 > 2x - 1
⇔ 4x + 8 > 2x - 1
⇔ 2x > - 9
⇔ x > \(\dfrac{-9}{2}\)
KL....
\(1+\dfrac{2x+1}{3}>\dfrac{2x-1}{6}\)
\(\Leftrightarrow\dfrac{6}{6}+\dfrac{2\left(2x+1\right)}{6}>\dfrac{2x-1}{6}\)
\(\Leftrightarrow6+4x+2>2x-1\)
\(\Leftrightarrow4x-2x>-1-6-2\)
\(\Leftrightarrow2x>-9\)
\(\Leftrightarrow x>-\dfrac{9}{2}\)
Vậy BPT có nghiệm \(x>-\dfrac{9}{2}\)
