Question

giải BPT :

\(1+\dfrac{2x+1}{3}>\dfrac{2x-1}{6}\)

Answer 1

\(1+\dfrac{2x+1}{3}>\dfrac{2x-1}{6}\)

\(\Leftrightarrow6+4x+2>2x-1\)

\(\Leftrightarrow4x-2x>-1-6-2\)

\(\Leftrightarrow x>-\dfrac{9}{2}\)

Vậy S = { x/ x > \(-\dfrac{9}{2}\)}

Answer 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....

Answer 3

\(1+\dfrac{2x+1}{3}>\dfrac{2x-1}{6}\)

\(\Leftrightarrow\dfrac{6}{6}+\dfrac{2\left(2x+1\right)}{6}>\dfrac{2x-1}{6}\)

\(\Leftrightarrow6+4x+2>2x-1\)

\(\Leftrightarrow4x-2x>-1-6-2\)

\(\Leftrightarrow2x>-9\)

\(\Leftrightarrow x>-\dfrac{9}{2}\)

Vậy BPT có nghiệm \(x>-\dfrac{9}{2}\)