a)\(\left[{}\begin{matrix}\dfrac{5}{2}-x=\dfrac{1}{3}\\\dfrac{5}{2}-x=-\dfrac{1}{3}\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{13}{6}\\x=\dfrac{17}{6}\end{matrix}\right.\)
b) 8/6-x-1/5=0
9/6-x=1/5
x=13/10
a) \(\left|2,5-x\right|=1,3\)
\(\Leftrightarrow\left[{}\begin{matrix}2,5-x=1,3\\2,5-x=-1,3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1,2\\x=3,8\end{matrix}\right.\)
b) \(1,6-\left|x-0,2\right|=0\)
\(\Leftrightarrow\left|x-0,2\right|=1,6\)
\(\Leftrightarrow\left[{}\begin{matrix}x-0,2=1,6\\x-0,2=-1,6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1,8\\x=-1,4\end{matrix}\right.\)
\(a,\left|2,5-x\right|=1,3\)
\(\Rightarrow2,5-x=\left\{{}\begin{matrix}-1,3\\1,3\end{matrix}\right.\)
\(TH1:2,5-x=1,3\)
\(x=2,5-1,3\)
\(x=1,2\)
\(TH2:2,5-x=-1,3\)
\(x=2,5-\left(-1,3\right)\)
\(x=3,8\)
Vậy \(x=\left\{{}\begin{matrix}1,2\\3,8\end{matrix}\right.\)
\(b,1,6-\left|x-0,2\right|=0\)
\(\left|x-0,2\right|=1,6-0=1,6\)
\(\Rightarrow x-0,2\) \(=\left\{{}\begin{matrix}-1,6\\1,6\end{matrix}\right.\)
\(TH1:x-0,2=-1,6\)
\(x=-1,6+0,2\)
\(x=-1,4\)
\(TH2:x-0,2=1,6\)
\(x=1,6+0,2\)
\(x=1,8\)
Vậy \(x\) \(=\left\{{}\begin{matrix}-1,4\\1,8\end{matrix}\right.\)