Câu 1:
\(\begin{array}{l}a)\,\frac{1}{6}.\left( {\frac{{ - 2}}{5}} \right) - \frac{1}{6}.\left( {\frac{{ - 6}}{5}} \right)\\ = \frac{1}{6}\left( {\frac{{ - 2}}{5} + \frac{6}{5}} \right)\\ = \frac{1}{6}.\frac{4}{5} = \frac{2}{{15}}\end{array}\)
\(b)\,\sqrt {64} - \sqrt {\frac{4}{{25}}} = 8 - \frac{2}{5} = \frac{{38}}{5}\)
Câu 2:
\(\begin{array}{l}a)\,\frac{{ - 3}}{{10}} + x = \frac{2}{5}\\ \Leftrightarrow x = \frac{2}{5} + \frac{3}{{10}}\\ \Leftrightarrow x = \frac{7}{{10}}\end{array}\)
\(\begin{array}{l}b)\,\left| {x - \frac{2}{5}} \right| - \frac{3}{4} = 1\\ \Leftrightarrow \left| {x - \frac{2}{5}} \right| = 1 + \frac{3}{4}\\ \Leftrightarrow \left| {x - \frac{2}{5}} \right| = \frac{7}{4}\,\,\,\,\left( * \right)\\TH1:\,x \ge \frac{2}{5}\\\left( * \right) \Leftrightarrow x - \frac{2}{5} = \frac{7}{4}\\ \Leftrightarrow x = \frac{7}{4} + \frac{2}{5}\\ \Leftrightarrow x = \frac{{35 + 8}}{{20}} = \frac{{43}}{{20}}\end{array}\)
TH2: \(x < \frac{2}{5}\)
\(\begin{array}{l}\left( * \right) \Leftrightarrow \,\,x - \frac{2}{5} = - \frac{7}{4}\\ \Leftrightarrow x = \frac{{ - 7}}{4} + \frac{2}{5}\\ \Leftrightarrow x = \frac{{ - 35 + 8}}{{20}}\\ \Leftrightarrow x = \frac{{27}}{{20}}\end{array}\)
