b) Ta có: \(\dfrac{x+1}{2}+2x\le\dfrac{x+5}{4}\)
\(\Leftrightarrow2\left(x+1\right)+8x\le x+5\)
\(\Leftrightarrow2x+2+8x-x-5\le0\)
\(\Leftrightarrow9x\le3\)
hay \(x\le\dfrac{1}{3}\)
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