Giải:
a) \(\dfrac{1}{3}.\left(-\dfrac{4}{5}\right)+\dfrac{1}{3}.\left(-\dfrac{6}{5}\right)\)
\(=\dfrac{1}{3}\left[-\dfrac{4}{5}+\left(-\dfrac{6}{5}\right)\right]\)
\(=\dfrac{1}{3}\left[-\dfrac{4}{5}-\dfrac{6}{5}\right]\)
\(=\dfrac{1}{3}\left(-\dfrac{10}{5}\right)\)
\(=\dfrac{-10}{15}=-\dfrac{2}{3}\)
Vậy ...
b) \(\dfrac{3}{7}.\dfrac{9}{25}-\dfrac{1}{14}.\dfrac{1}{13}\)
\(=\dfrac{1}{7}.\dfrac{27}{25}-\dfrac{1}{7}.\dfrac{1}{26}\)
\(=\dfrac{1}{7}\left(\dfrac{27}{25}-\dfrac{1}{26}\right)\)
\(=\dfrac{1}{7}.\dfrac{677}{650}\)
\(=\dfrac{677}{4550}\)
Vậy ...
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