Bài 2 (6.45). Tính một cách hợp lí:
a) \(A = \frac{{ - 3}}{{14}} + \frac{2}{{13}} + \frac{{ - 25}}{{14}} + \frac{{ - 15}}{{13}}\)
b) \(B = \frac{5}{3}.\frac{7}{{25}} + \frac{5}{3}.\frac{{21}}{{25}} - \frac{5}{3}.\frac{7}{{25}}\)
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a) \(A \) \(= \frac{{ - 3}}{{14}} + \frac{2}{{13}} + \frac{{ - 25}}{{14}} + \frac{{ - 15}}{{13}} \) \(= \left( {\frac{{ - 3}}{{14}} + \frac{{ - 25}}{{14}}} \right) + \left( {\frac{2}{{13}} + \frac{{ - 15}}{{13}}} \right)\\ \) \(= \frac{{ - 28}}{{14}} + \frac{{ - 13}}{{13}} \) \(= - 2 - 1 \) \(= - 3\)
b) \(B \) \(= \frac{5}{3}.\frac{7}{{25}} + \frac{5}{3}.\frac{{21}}{{25}} - \frac{5}{3}.\frac{7}{{25}} \) \(= \frac{5}{3}.\left( {\frac{7}{{25}} + \frac{{21}}{{25}} - \frac{7}{{25}}} \right)\\ \) \(= \frac{5}{3}.\left( {\frac{7}{{25}} - \frac{7}{{25}} + \frac{{21}}{{25}}} \right) \) \(= \frac{5}{3}.\frac{{21}}{{25}} \) \(= \frac{7}{5}\)