Ta có : \(15\) - \(\dfrac{4x}{3}\) ≥ 5
⇔ \(\dfrac{15.3}{3}\) - \(\dfrac{4x}{3}\) ≥ \(\dfrac{5.3}{3}\)
⇒ 15.3 - 4x ≥ 5.3
⇔ 45 - 4x ≥ 15
⇔ - 4x ≥ 15 - 45
⇔ - 4x ≥ -30
⇔ x ≤ \(\dfrac{-30}{-4}\)
⇔ x ≤ \(\dfrac{15}{2}\)
Vậy phương trình \(15\) - \(\dfrac{4x}{3}\) lớn hơn hoặc bằng 5 có S = { \(\dfrac{15}{2}\) }
\(\dfrac{15-4x}{3}\ge5\)
\(\Leftrightarrow\dfrac{15-4x}{3}\ge\dfrac{15}{3}\)
\(\Leftrightarrow15-4x\ge15\)
\(\Leftrightarrow-4x\ge0\)
\(\Leftrightarrow x\le0\)
