1) Ta có: \(3x\left(2-5x\right)+35x-14=0\)
\(\Leftrightarrow3x\left(2-5x\right)+7\left(5x-2\right)=0\)
\(\Leftrightarrow-3x\left(5x-2\right)+7\left(5x-2\right)=0\)
\(\Leftrightarrow\left(5x-2\right)\left(-3x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-2=0\\-3x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=2\\-3x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=\dfrac{7}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{2}{5};\dfrac{7}{3}\right\}\)
2) Ta có: \(4x-6+5x\left(3-2x\right)=0\)
\(\Leftrightarrow2\left(2x-3\right)-5x\left(2x-3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2-5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\5x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{2}{5}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{3}{2};\dfrac{2}{5}\right\}\)
