`a.` \(\dfrac{6x+5}{2}-\dfrac{10x+3}{4}\ge2x+\dfrac{2x+1}{2}\)
\(\Leftrightarrow\dfrac{12x+10-10x-3}{4}\ge\dfrac{4x+2x+1}{2}\)
\(\Leftrightarrow\dfrac{2x+7}{4}\ge\dfrac{6x+1}{2}\)
\(\Leftrightarrow\dfrac{2x+7}{4}\ge\dfrac{12x+2}{4}\)
\(\Leftrightarrow2x+7\ge12x+2\)
\(\Leftrightarrow-10x\ge-5\)
\(\Leftrightarrow x\le\dfrac{1}{2}\)
Vậy \(S=\left\{x|x\le\dfrac{1}{2}\right\}\)
`b.`
Ta có:
\(x+y=3\)
\(\Leftrightarrow\dfrac{x}{2}+\dfrac{x}{2}+y=3\)
Áp dụng BĐT AM-GM, ta có:
\(\dfrac{x}{2}+\dfrac{x}{2}+y\ge3\sqrt[3]{\dfrac{x.x.y}{2.2}}\)
\(\Leftrightarrow\dfrac{x}{2}+\dfrac{x}{2}+y\ge3\sqrt[3]{\dfrac{x^2y}{4}}\)
\(\Leftrightarrow3\ge3\sqrt[3]{\dfrac{x^2y}{4}}\)
\(\Leftrightarrow1\ge\sqrt[3]{\dfrac{x^2y}{4}}\)
\(\Leftrightarrow1\ge\dfrac{x^2y}{4}\)
\(\Leftrightarrow4\ge x^2y\) hay \(x^2y\le4\left(đfcm\right)\)
