a) \( |x -1,7| = 2,3\)
\(\eqalign{
& \Rightarrow \left[ \matrix{
x - 1,7 = 2,3 \hfill \cr
x - 1,7 = - 2,3 \hfill \cr} \right. \cr
& \Rightarrow \left[ \matrix{
x = 2,3 + 1,7 \hfill \cr
x = - 2,3 + 1,7 \hfill \cr} \right. \cr
& \Rightarrow \left[ \matrix{
x = 4 \hfill \cr
x = - 0,6 \hfill \cr} \right. \cr} \)
Vậy \(x = 4\) hoặc \(x = -0,6\)
b)
\(\begin{array}{l}
\left| {x + \dfrac{3}{4}} \right| - \dfrac{1}{3} = 0\\
\left| {x + \dfrac{3}{4}} \right| = \dfrac{1}{3}\\
\text{Trường hợp 1}:\\x + \dfrac{3}{4} = \dfrac{1}{3}\\x = \dfrac{1}{3} - \dfrac{3}{4}\\
x = \dfrac{{ - 5}}{{12}}\\
\text{Trường hợp 2}:\\x + \dfrac{3}{4} = - \dfrac{1}{3}\\x = - \dfrac{1}{3} - \dfrac{3}{4}\\
x = \dfrac{{ - 13}}{{12}}
\end{array}\)
Vậy \( x = \dfrac{{ - 5}}{{12}}\) hoặc \({x = \dfrac{{ - 13}}{{12}}}\).
