Câu 1. (2 điểm) Tính.
a) $\dfrac{1}{3}+\dfrac{3}{4}-\dfrac{5}{6}$ ;
b) $\dfrac{-2}{3}+\dfrac{6}{5}:\dfrac{2}{3}-\dfrac{2}{15}$ ;
c) $\dfrac{-3}{7}+\dfrac{5}{13}+\dfrac{-4}{7}$ ;
d) $\dfrac{12}{19}+\dfrac{-8}{13}-\dfrac{12}{19}+\dfrac{5}{-13}+2$ .
a) \(\frac{1}{3}+\frac{3}{4}-\frac{5}{6}\)
\(=\frac{1}{3}+\frac{3}{4}+\frac{-5}{6}\)
\(=\frac{1.4+3.3+\left(-5\right).2}{12}\)
\(=\frac{4+9+\left(-10\right)}{12}\)
\(=\frac{3}{12}\)
\(=\frac{1}{4}\)
b) \(\frac{-2}{3}+\frac{6}{5}\div\frac{2}{3}-\frac{2}{15}\text{ }\)
\(=\frac{-2}{3}+\frac{6}{5}\times\frac{3}{2}-\frac{2}{15}\)
\(=\frac{-2}{3}+\frac{18}{10}-\frac{2}{15}\)
\(=\frac{-2}{3}+\frac{9}{5}+\frac{-2}{15}\)
\(=\frac{\left(-2\right).5+9.3+\left(-2\right)}{15}\)
\(=\frac{\left(-10\right)+27+\left(-2\right)}{15}\)
\(=\frac{15}{15}\)
\(=1\)