Trả lời câu hỏi 4 Bài 9 trang 33 Toán 6 Tập 2

Đề bài

Tính \(\dfrac{3}{5} - \dfrac{{ - 1}}{2}\); \(\dfrac{{ - 5}}{7} - \dfrac{1}{3}\); \(\dfrac{{ - 2}}{5} - \dfrac{{ - 3}}{4}\); \( - 5 - \dfrac{1}{6}\) 

Lời giải chi tiết

Ta có 

+) \(\dfrac{3}{5} - \dfrac{{ - 1}}{2} = \dfrac{3}{5} + \dfrac{1}{2} \)\(= \dfrac{6}{{10}} + \dfrac{5}{{10}} = \dfrac{{11}}{{10}}\)

+) \(\dfrac{{ - 5}}{7} - \dfrac{1}{3} = \dfrac{{ - 5}}{7} + \left( {\dfrac{{ - 1}}{3}} \right) = \dfrac{{ - 15}}{{21}} + \dfrac{{ - 7}}{{21}} \)\(= \dfrac{{ - 15 + \left( { - 7} \right)}}{{21}} = \dfrac{{ - 22}}{{21}}\)

+) \(\dfrac{{ - 2}}{5} - \dfrac{{ - 3}}{4} = \dfrac{{ - 2}}{5} + \dfrac{3}{4} \)\(= \dfrac{{ - 8}}{{20}} + \dfrac{{15}}{{20}} = \dfrac{{ - 8 + 15}}{{20}} = \dfrac{7}{{20}}\)

+) \( - 5 - \dfrac{1}{6} =  - 5 + \left( {\dfrac{{ - 1}}{6}} \right) = \dfrac{{ - 30}}{6} + \dfrac{{ - 1}}{6} \)\(= \dfrac{{ - 30 + \left( { - 1} \right)}}{6} = \dfrac{{ - 31}}{6}\)