Question

giải bất phương trình sau \(\dfrac{x}{3}+\dfrac{2x-4}{4}\ge\dfrac{x}{6}+x\)

Answer 1

\(\dfrac{x}{3}+\dfrac{2x-4}{4}\ge\dfrac{x}{6}+x\)

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

\(\Leftrightarrow\dfrac{2x}{6}-\dfrac{x}{6}+\dfrac{x}{2}-\dfrac{2x}{2}\ge1\)

\(\Leftrightarrow\dfrac{x}{6}-\dfrac{x}{2}\ge1\)

\(\Leftrightarrow\dfrac{x-3x}{6}\ge1\)

\(\Leftrightarrow-2x\ge6\)

\(\Leftrightarrow x\ge-3\)

Vậy BPT có tập nghiệm là ;\(S=\left\{x\ge-3\right\}\)