Ta có : \(\dfrac{x+5}{2}+\dfrac{3-2x}{4}=x-\dfrac{x+7}{6}\)
\(\Leftrightarrow6\left(x+5\right)+3\left(3-2x\right)=12x-2\left(x+7\right)\)
\(\Leftrightarrow6x+30+9-6x=12-2x-14\)
\(\Leftrightarrow6x+30+9-6x-12+2x+14=0\)
\(\Leftrightarrow2x+41=0\)
\(\Leftrightarrow x=-\dfrac{41}{2}\)
Vậy ...
\(\Leftrightarrow-10x+53=0\)
\(\Leftrightarrow x=\dfrac{53}{10}\)
Vậy ...
( Lỗi sửa lại :vvv )
Ta có: \(\dfrac{x+5}{2}+\dfrac{3-2x}{4}=x-\dfrac{7+x}{6}\)
\(\Leftrightarrow\dfrac{6\left(x+5\right)}{12}+\dfrac{3\left(3-2x\right)}{12}=\dfrac{12x}{12}-\dfrac{2\left(7+x\right)}{12}\)
\(\Leftrightarrow6x+30+9-6x=12x-14-2x\)
\(\Leftrightarrow10x-14=39\)
\(\Leftrightarrow10x=53\)
\(\Leftrightarrow x=\dfrac{53}{10}\)
Vậy: \(S=\left\{\dfrac{53}{10}\right\}\)
