\(a,\Rightarrow x-\dfrac{1}{8}\cdot\dfrac{-12}{16}=\dfrac{2}{5}\\ \Rightarrow x+\dfrac{3}{32}=\dfrac{2}{5}\\ \Rightarrow x=\dfrac{2}{5}-\dfrac{3}{32}=\dfrac{49}{160}\)
\(d,\left(x+\dfrac{1}{4}\right)\left(x-\dfrac{3}{7}\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)
\(x-\dfrac{1}{8}:\dfrac{-16}{12}=\dfrac{2}{5}\)
\(x+\dfrac{3}{32}=\dfrac{2}{5}\)
\(x=\dfrac{49}{160}\)
\(x-\dfrac{1}{8}:\dfrac{-16}{12}=\dfrac{2}{5}\)
\(\Rightarrow x-\dfrac{1}{8}.\dfrac{-12}{16}=\dfrac{2}{5}\)
\(\Rightarrow x-\dfrac{1}{8}.\dfrac{-3}{4}=\dfrac{2}{5}\)
\(\Rightarrow x+\dfrac{3}{32}=\dfrac{2}{5}\)
\(\Rightarrow x=\dfrac{49}{160}\)
