Đề bài

Bỏ ngoặc rồi tính a)\(\left( {\dfrac{{ - 3}}{8}} \right) + \left( {\dfrac{7}{9} - \dfrac{5}{8}} \right)\) b)\(\dfrac{4}{9} - \left( {\dfrac{3}{7} + \dfrac{2}{9}} \right)\) c)\(\left[ {\left( {\dfrac{{ - 2}}{5}} \right) + \dfrac{1}{3}} \right] - \left( {\dfrac{3}{5} - \dfrac{1}{4}} \right)\) d)\(\left( {1\dfrac{1}{2} - \dfrac{3}{4}} \right) - \left( {0,25 + \dfrac{1}{2}} \right)\) 

Lời giải chi tiết

a) \(\left( {\dfrac{{ - 3}}{8}} \right) + \left( {\dfrac{7}{9} - \dfrac{5}{8}} \right)\) \( = \left( {\dfrac{{ - 3}}{8}} \right) - \dfrac{5}{8} + \dfrac{7}{9} \\=  - 1 + \dfrac{7}{9} \\= \dfrac{{ - 9}}{9} + \dfrac{7}{9} \\= \dfrac{{ - 2}}{9}\) b) \(\dfrac{4}{9} - \left( {\dfrac{3}{7} + \dfrac{2}{9}} \right)\) \(=\dfrac{4}{9} - \dfrac{3}{7} - \dfrac{2}{9} \\= \dfrac{4}{9} - \dfrac{2}{9} - \dfrac{3}{7} \\= \dfrac{2}{9} - \dfrac{3}{7} \\= \dfrac{{14}}{{63}} - \dfrac{{27}}{{63}} \\= \dfrac{{ - 13}}{{63}}\) c)\(\left[ {\left( {\dfrac{{ - 2}}{5}} \right) + \dfrac{1}{3}} \right] - \left( {\dfrac{3}{5} - \dfrac{1}{4}} \right)\) \(\begin{array}{l} = \left( {\dfrac{{ - 2}}{5}} \right) + \dfrac{1}{3} - \dfrac{3}{5} + \dfrac{1}{4}\\ = \left( {\dfrac{{ - 2}}{5}} \right) - \dfrac{3}{5} + \dfrac{1}{4} + \dfrac{1}{3}\\ = \left( { - 1} \right) + \dfrac{1}{4} + \dfrac{1}{3}\\ = \left( {\dfrac{{ - 12}}{{12}}} \right) + \dfrac{3}{{12}} + \dfrac{4}{{12}}\\ = \dfrac{{ - 5}}{{12}}\end{array}\) d)\(\left( {1\dfrac{1}{2} - \dfrac{3}{4}} \right) - \left( {0,25 + \dfrac{1}{2}} \right)\) \(=\left( {\dfrac{3}{2} - \dfrac{3}{4}} \right) - \left( {\dfrac{1}{4} + \dfrac{1}{2}} \right) \\= \left( {\dfrac{3}{2} - \dfrac{1}{2}} \right) - \left( {\dfrac{3}{4} + \dfrac{1}{4}} \right) \\= 1 - 1 = 0\)