c: Ta có: \(\dfrac{3}{2}+\dfrac{4}{5}\left|x-\dfrac{3}{4}\right|=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{4}{5}\cdot\left|x-\dfrac{3}{4}\right|=\dfrac{7}{4}-\dfrac{3}{2}=\dfrac{1}{4}\)
\(\Leftrightarrow\left|x-\dfrac{3}{4}\right|=\dfrac{1}{4}:\dfrac{4}{5}=\dfrac{5}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{5}{16}\\x-\dfrac{3}{4}=-\dfrac{5}{16}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{16}\\x=\dfrac{7}{16}\end{matrix}\right.\)
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