Lời giải:
a)
\(||x+3|-8|=20\)
\(\Rightarrow \left[\begin{matrix} |x+3|-8=20\\ |x+3|-8=-20\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} |x+3|=28\\ |x+3|=-12(\text{vô lý-loại})\end{matrix}\right.\)
\( \Rightarrow \left[\begin{matrix} x+3=28\\ x+3=-28\end{matrix}\right.\) \(\Rightarrow \left[\begin{matrix} x=25\\ x=-31\end{matrix}\right.\)
Vậy..........
b)
Nếu \(x\geq 7\Rightarrow \left\{\begin{matrix} |x-7|=x-7\\ |x+\frac{2}{3}|=x+\frac{2}{3}\end{matrix}\right.\)
PT trở thành: \(x+\frac{2}{3}-(x-7)=\frac{5}{3}\)
\(\Leftrightarrow \frac{23}{3}=\frac{5}{3}\) (vô lý- loại)
Nếu \(\frac{-2}{3}\le x< 7\Rightarrow \left\{\begin{matrix} |x+\frac{2}{3}|=x+\frac{2}{3}\\ |x-7|=7-x\end{matrix}\right.\)
PT trở thành: \(x+\frac{2}{3}-(7-x)=\frac{5}{3}\Leftrightarrow x=4\) (thỏa mãn)
Nếu \(x< \frac{-2}{3}\Rightarrow \left\{\begin{matrix} |x+\frac{2}{3}=-(x+\frac{2}{3})\\ |x-7|=7-x\end{matrix}\right.\)
PT trở thành: \(-(x+\frac{2}{3})-(7-x)=\frac{5}{3}\Leftrightarrow \frac{-23}{3}=\frac{5}{3}\) (vô lý- loại)
Vậy $x=4$
