a: \(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=2\\x+\dfrac{1}{5}=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{5}\\x=-\dfrac{11}{5}\end{matrix}\right.\)
\(a,\left|x+\dfrac{1}{5}\right|-4=-2\\ \left|x+\dfrac{1}{5}\right|=\left(-2\right)+4\\ \left|x+\dfrac{1}{5}\right|=2\Rightarrow x+\dfrac{1}{5}=2\)
hoặc \(x+\dfrac{1}{5}=-2\)
Với \(x+\dfrac{1}{5}=2\Rightarrow x=2-\dfrac{1}{5}\)hay \(x=\dfrac{9}{5}\)
Với \(x+\dfrac{1}{5}=-2\Rightarrow x=-2-\dfrac{1}{5}\)hay \(x=-\dfrac{11}{5}\)
Thông cảm . bt làm mỗi câu a
a/ \(\left|x+\dfrac{1}{5}\right|-4=-2\)
\(\left|x+\dfrac{1}{5}\right|=-2+4\)
\(\left|x+\dfrac{1}{5}\right|=2\)
\(x+\dfrac{1}{5}=2\) hoặc \(x+\dfrac{1}{5}=-2\)
\(x=2-\dfrac{1}{5}\) hoặc \(x+\dfrac{1}{5}=-2-\dfrac{1}{5}\)
\(x=\dfrac{10}{5}-\dfrac{1}{5}\) hoặc \(x=\dfrac{-10}{5}-\dfrac{1}{5}\)
\(x=\dfrac{9}{5}\) hoặc \(x=\dfrac{-11}{5}\)
