Ta có: \(\dfrac{5\left(1-2x\right)}{3}+\dfrac{x}{2}=\dfrac{3\left(x-5\right)}{4}\)
\(\Leftrightarrow\dfrac{20\left(1-2x\right)}{12}+\dfrac{6x}{12}=\dfrac{9\left(x-5\right)}{12}\)
\(\Leftrightarrow20-4x+6x=9x-45\)
\(\Leftrightarrow2x+20-9x+45=0\)
\(\Leftrightarrow-7x+65=0\)
\(\Leftrightarrow-7x=-65\)
hay \(x=\dfrac{65}{7}\)
Vậy: \(S=\left\{\dfrac{65}{7}\right\}\)
\(\dfrac{5\left(1-2x\right)}{3}+\dfrac{x}{2}=\dfrac{3\left(x-5\right)}{4}\)
\(\Leftrightarrow\dfrac{20-40x}{12}+\dfrac{6x}{12}=\dfrac{9x-45}{12}\)
\(\Rightarrow20-40x+6x=9x-45\)
\(\Leftrightarrow6x-40x-9x=-45-20\)
\(-43x=-65\)
\(\Leftrightarrow x=\dfrac{65}{43}\)
