Giải Bài 1 trang 17 SBT toán 7 tập 1 – Chân trời sáng tạo: Bỏ ngoặc rồi tính a) – 3/8 + 7/9 – 5/8 b) 4/9 – 3/7 + 2/9

Đề bài/câu hỏi:

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

Hướng dẫn:

Lý thuyết Quy tắc dấu ngoặc và quy tắc chuyển vế SGK Toán 7 ()

Lời giải:

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