\(\Leftrightarrow\dfrac{6\left(x-1\right)}{12}-\dfrac{4\left(x-2\right)}{12}-\dfrac{12x}{12}+\dfrac{3\left(x-3\right)}{12}\le0\)
\(\Leftrightarrow\dfrac{6x-6-4x+8-12x+3x-9}{12}\le0\)
\(\Leftrightarrow-7x-7\le0\)
\(\Leftrightarrow-7x\le7\)
\(\Leftrightarrow x\ge-1\)
\(\Leftrightarrow\dfrac{3\left(x-1\right)-2\left(x-2\right)}{6}\le\dfrac{4x-\left(x-3\right)}{4}\)
\(\Leftrightarrow\dfrac{3x-3-2x+4}{6}\le\dfrac{4x-x+3}{4}\)
\(\Leftrightarrow\dfrac{x+1}{6}\le\dfrac{3x+3}{4}\)
\(\Leftrightarrow\dfrac{x+1}{6}-\dfrac{3\left(x+1\right)}{4}\le0\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{6}-\dfrac{3}{4}\right)\le0\)
\(\Leftrightarrow\left(x+1\right).-\dfrac{7}{12}\le0\)
\(\Leftrightarrow x+1\ge0\)
\(\Leftrightarrow x\ge-1\)
Vậy \(S=\left\{x|x\ge-1\right\}\)
\(\dfrac{6\left(x-1\right)}{12}-\dfrac{4\left(x-2\right)}{12}\le\dfrac{12x}{12}-\dfrac{3\left(x-3\right)}{12}\\ =>6x-6-4x+8\le12x-3x+9\\ < =>2x+2\le9x+9\\ < =>2x-9x\le9-2\\ 7x\le7< =>x\le1\)
vậy pt có nghiệm \(x\le1\)
