Bài 1 trang 17 SBT Toán 7 tập 1 Chân trời sáng tạo: Bỏ ngoặc rồi tính...

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)$\left( {1\dfrac{1}{2} - \dfrac{3}{4}} \right) - \left( {0,25 + \dfrac{1}{2}} \right)$ 

Lý thuyết Quy tắc dấu ngoặc và quy tắc chuyển vế SGK Toán 7 Chân trời sáng tạo (baitapsgk.com)

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)$\left( {1\dfrac{1}{2} - \dfrac{3}{4}} \right) - \left( {0,25 + \dfrac{1}{2}} \right)$=$\left( {\dfrac{3}{2} - \dfrac{3}{4}} \right) - \left( {\dfrac{1}{4} + \dfrac{1}{2}} \right) = \left( {\dfrac{3}{2} - \dfrac{1}{2}} \right) - \left( {\dfrac{3}{4} + \dfrac{1}{4}} \right) = 1 - 1 = 0$