a) \(\left|3x+1\right|=2-\left|-\dfrac{4}{5}\right|\)
\(\left|3x+1\right|=2-\dfrac{4}{5}\)
\(\left|3x+1\right|=\dfrac{6}{5}\)
TH1: \(3x+1=-\dfrac{6}{5}\)
\(3x=-\dfrac{6}{5}-1\)
\(3x=\dfrac{-11}{5}\)
\(x=\dfrac{-11}{5}\div3\)
\(x=\dfrac{-11}{15}\)
TH2: \(3x+1=\dfrac{6}{5}\)
\(3x=\dfrac{6}{5}-1\)
\(3x=\dfrac{1}{5}\)
\(x=\dfrac{1}{5}\div3\)
\(x=\dfrac{1}{15}\)
Câu b tương tự.