Giải bài 2 (6.45) trang 24 vở thực hành Toán 6 Q2

Đề bài

Bài 2 (6.45). Tính một cách hợp lí:

a) \(A = \frac{{ – 3}}{{14}} + \frac{2}{{13}} + \frac{{ – 25}}{{14}} + \frac{{ – 15}}{{13}}\)

b) \(B = \frac{5}{3}.\frac{7}{{25}} + \frac{5}{3}.\frac{{21}}{{25}} – \frac{5}{3}.\frac{7}{{25}}\)

Lời giải chi tiết

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) \(B \) \(= \frac{5}{3}.\frac{7}{{25}} + \frac{5}{3}.\frac{{21}}{{25}} – \frac{5}{3}.\frac{7}{{25}} \) \(= \frac{5}{3}.\left( {\frac{7}{{25}} + \frac{{21}}{{25}} – \frac{7}{{25}}} \right)\\ \) \(= \frac{5}{3}.\left( {\frac{7}{{25}} – \frac{7}{{25}} + \frac{{21}}{{25}}} \right) \) \(= \frac{5}{3}.\frac{{21}}{{25}} \) \(= \frac{7}{5}\)