a) Ta có: \(\frac{3}{4}-x=\frac{1}{5}\)
hay \(x=\frac{3}{4}-\frac{1}{5}=\frac{11}{20}\)
Vậy: \(x=\frac{11}{20}\)
b) Ta có: \(\left|x+\frac{2}{5}\right|-\frac{3}{7}=\frac{4}{7}\)
\(\Leftrightarrow\left|x+\frac{2}{5}\right|=\frac{4}{7}+\frac{3}{7}=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{2}{5}=1\\x+\frac{2}{5}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1-\frac{2}{5}=\frac{3}{5}\\x=-1-\frac{2}{5}=\frac{-7}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{3}{5};\frac{-7}{5}\right\}\)
c) Ta có: \(\left(x+\frac{1^3}{3}\right):2=\frac{-1}{16}\)
\(\Leftrightarrow x+\frac{1}{3}=\frac{-1}{16}\cdot2=-\frac{1}{8}\)
hay \(x=\frac{-1}{8}-\frac{1}{3}=-\frac{11}{24}\)
Vậy: \(x=\frac{-11}{24}\)
d) Ta có: \(\frac{x+2}{3}=\frac{12}{x+2}\)
\(\Leftrightarrow\left(x+2\right)^2=36\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=6\\x+2=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-8\end{matrix}\right.\)(tm)
Vậy: \(x\in\left\{4;-8\right\}\)
