a, \(\dfrac{-3x}{4}.\left(\dfrac{3}{x}+\dfrac{5}{7}\right)=0\) \(\left(x\ne0\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{-3x}{4}=0\\\dfrac{3}{x}+\dfrac{5}{7}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-4,2\left(tm\right)\end{matrix}\right.\)
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b, \(x+\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{4}\)
\(\Leftrightarrow x=\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{2}\)
\(\Leftrightarrow x=\dfrac{1}{12}\)
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