h: Ta có: \(\dfrac{2-x}{4}=\dfrac{3x-1}{-3}\)
\(\Leftrightarrow-3\left(2-x\right)=4\left(3x-1\right)\)
\(\Leftrightarrow-6+3x=12x-4\)
\(\Leftrightarrow-9x=2\)
hay \(x=-\dfrac{2}{9}\)
i: Ta có: \(\dfrac{12-3x}{32}=\dfrac{6}{4-x}\)
\(\Leftrightarrow3\left(4-x\right)^2=192\)
\(\Leftrightarrow\left(x-4\right)^2=64\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=8\\x-4=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-4\end{matrix}\right.\)
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