\(a,\dfrac{4-2x}{-7}+3< \dfrac{x+1}{3}\)
\(\Leftrightarrow\dfrac{-2+2x}{7}+3< \dfrac{x+1}{3}\)
\(\Leftrightarrow3\left(-2+2x\right)+3< 7\left(x+1\right)\)
\(\Leftrightarrow-6+6x+3< 7x+7\)
\(\Leftrightarrow6x-7x< 7-3+6\)
\(\Leftrightarrow-x< 10\)
\(\Leftrightarrow x>-10\)
Vậy \(S=\left\{x|x>-10\right\}\)
\(b,\left(x-1\right)\left(-x+5\right)\ge\left(2-x\right)\left(2+x\right)\)
\(\Leftrightarrow-x^2+5x+x-5\ge4-x^2\)
\(\Leftrightarrow-x^2+x^2+5x+x-4-5\ge0\)
\(\Leftrightarrow6x-9\ge0\)
\(\Leftrightarrow6x\ge9\)
\(\Leftrightarrow x\ge\dfrac{3}{2}\)
Vậy \(S=\left\{x|x\ge\dfrac{3}{2}\right\}\)
\(c,\dfrac{5-2x}{-2}+4\le\dfrac{3-x}{5}-\dfrac{1}{2}\)
\(\Leftrightarrow-5\left(5-2x\right)+4\le2\left(3-x\right)-5\)
\(\Leftrightarrow-25+10x+4\le6-2x-5\)
\(\Leftrightarrow10x+2x\le6-5+25-4\)
\(\Leftrightarrow12x\le22\)
\(\Leftrightarrow x\le\dfrac{11}{6}\)
Vậy \(S=\left\{\dfrac{11}{6}\right\}\)
\(d,\left(x-1\right)\left(x+2\right)\le-3x-6\)
\(\Leftrightarrow x^2+2x-x-2\le-3x-6\)
\(\Leftrightarrow x^2+2x-x+3x-2+6\le0\)
\(\Leftrightarrow x^2+4x+4\le0\)
\(\Leftrightarrow\left(x+2\right)^2\le0\)
\(\Leftrightarrow x\le-2\)
Vậy \(S=\left\{x|x\le-2\right\}\)
