\(2\left(|x-1|+x-\frac{4}{5}\right)=2x-\frac{2}{5}\)

\(TH1:x\ge1\Rightarrow|x-1|=x-1\)

\(\Rightarrow2\left(x-1+x-\frac{4}{5}\right)=2x-\frac{2}{5}\)

\(\Rightarrow2\left(2x-\frac{9}{5}\right)=2x-\frac{2}{5}\Rightarrow4x-\frac{18}{5}=2x-\frac{2}{5}\)

\(\Rightarrow4x-2x=\frac{18}{5}-\frac{2}{5}\Rightarrow2x=\frac{16}{5}\Rightarrow x=\frac{16}{5}:2=\frac{16}{10}=\frac{8}{5}\)

\(TH2:x< 1\Rightarrow|x-1|=-x+1\)

\(\Rightarrow2\left(-x+1+x-\frac{4}{5}\right)=2x-\frac{2}{5}\)

\(\Rightarrow2\left(1-\frac{4}{5}\right)=2x-\frac{2}{5}\Rightarrow2\cdot\frac{1}{5}=2x-\frac{2}{5}\)

\(\Rightarrow2x-\frac{2}{5}=\frac{2}{5}\Rightarrow2x=\frac{2}{5}+\frac{2}{5}=\frac{4}{5}\Rightarrow x=\frac{4}{5}:2=\frac{4}{10}=\frac{2}{5}\)

a.x=4

giai chi tiet gium minh voi

Ta có : 

\(4-\left|x-\frac{1}{5}\right|=-\frac{1}{2}\)

\(\Leftrightarrow\)\(\left|x-\frac{1}{5}\right|=4-\left(-\frac{1}{2}\right)\)

\(\Leftrightarrow\)\(\left|x-\frac{1}{5}\right|=\frac{9}{2}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-\frac{1}{5}=\frac{9}{2}\\x-\frac{1}{5}=-\frac{9}{2}\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{9}{2}+\frac{1}{5}\\x=-\frac{9}{2}+\frac{1}{5}\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{43}{10}\\x=-\frac{43}{10}\end{cases}}\)

 Vậy \(x=\frac{43}{10}\) hoặc \(x=-\frac{43}{10}\)

Ta có: \(4-\left|x-\frac{1}{5}\right|=\frac{-1}{2}\)

\(\Rightarrow\left|x-\frac{1}{5}\right|=4-\frac{-1}{2}\)

\(\Rightarrow\left|x-\frac{1}{5}\right|=\frac{9}{2}\)

\(\Rightarrow x-\frac{1}{5}=\pm\frac{9}{2}\)

Nếu \(x-\frac{1}{5}=\frac{9}{2}\Rightarrow x=\frac{9}{2}+\frac{1}{5}=\frac{47}{10}\)

Nếu \(x-\frac{1}{5}=\frac{-9}{2}\Rightarrow x=\frac{-9}{2}+\frac{1}{5}=\frac{-43}{10}\)

Vậy \(x=\left\{\frac{47}{10};\frac{-43}{10}\right\}\)

Chúc bạn năm mới vui vẻ!