a) \(\dfrac{-7}{12}-\left(\dfrac{3}{5}+x\right)=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{-7}{12}-\dfrac{3}{5}-x=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{-71}{60}-x=\dfrac{3}{4}\)
\(\Leftrightarrow x=\dfrac{-71}{60}-\dfrac{3}{4}\)
\(\Leftrightarrow x=\dfrac{-29}{15}\)
Vậy \(x=\dfrac{-29}{15}\)
b) \(2017x\left(x-\dfrac{2006}{7}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2017x=0\\x-\dfrac{2006}{7}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2006}{7}\end{matrix}\right.\)
Vậy \(x=0\) ; \(x=\dfrac{2006}{7}\)
c) \(5\left(x-2\right)+3x\left(2-x\right)=0\)
\(\Leftrightarrow5\left(x-2\right)-3x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(5-3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\5-3x=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\)
Vậy \(x=2\) ; \(x=\dfrac{5}{3}\)
