Giải Bài 1 trang 19 SBT toán 7 tập 1 – Chân trời sáng tạo: Thực hiện phép tính: a) 3/7 + 3/7: – 3/2 – 1/2 b) 21/2 – – 1/2 ^2 + 3/4 c) 3/8 – 1, 25 : 3/4 – 0

Đề bài/câu hỏi:

Thực hiện phép tính:

\(a)\dfrac{3}{7} + \dfrac{3}{7}:\left( { – \dfrac{3}{2}} \right) – \dfrac{1}{2}\)

\(b)\,\,2\dfrac{1}{2} – {\left( { – \dfrac{1}{2}} \right)^2} + \dfrac{3}{4}\)

\(c)\left( {\dfrac{3}{8} – 1,25} \right):\left( {\dfrac{3}{4} – 0,25} \right)\)

\(d)\,\,1\dfrac{2}{5}:\dfrac{{14}}{{15}} + \left( {1\dfrac{1}{3} – 2\dfrac{1}{2}} \right):\dfrac{5}{6}\)

Hướng dẫn:

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

Lời giải:

\(\begin{array}{l}a)\dfrac{3}{7} + \dfrac{3}{7}:\left( { – \dfrac{3}{2}} \right) – \dfrac{1}{2} \\= \dfrac{3}{7} + \dfrac{3}{7}.\left( { – \dfrac{2}{3}} \right) – \dfrac{1}{2}\\ = \dfrac{3}{7}.\left( {1 – \dfrac{2}{3}} \right) – \dfrac{1}{2} = \dfrac{3}{7}.\dfrac{1}{3} – \dfrac{1}{2}\\ = \dfrac{1}{7} – \dfrac{1}{2}= \dfrac{2}{14} – \dfrac{7}{14} = – \dfrac{5}{{14}}\\b)\,\,2\dfrac{1}{2} – {\left( { – \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} = \dfrac{5}{2} – \dfrac{1}{4} + \dfrac{3}{4}\\ = \dfrac{{10}}{4} – \dfrac{1}{4} + \dfrac{3}{4} = \dfrac{{12}}{4} = 3\\c)\left( {\dfrac{3}{8} – 1,25} \right):\left( {\dfrac{3}{4} – 0,25} \right) \\= \left( {\dfrac{3}{8} – \dfrac{5}{4}} \right):{\left( {\dfrac{3}{4} – \dfrac{1}{4}} \right)^2}\\ = \left( {\dfrac{{ – 7}}{8}} \right):{\left( {\dfrac{1}{2}} \right)^2}\\ = \dfrac{{ – 7}}{8}:\dfrac{1}{4} = \dfrac{{ – 7}}{8}.4 = \dfrac{{ – 7}}{2}\\d)\,\,1\dfrac{2}{5}:\dfrac{{14}}{{15}} + \left( {1\dfrac{1}{3} – 2\dfrac{1}{2}} \right):\dfrac{5}{6}\\ = \dfrac{7}{5}:\dfrac{{14}}{{15}} + \left( {\dfrac{4}{3} – \dfrac{5}{2}} \right):\dfrac{5}{6}\\ = \dfrac{7}{5}.\dfrac{{15}}{{14}} + \left( { – \dfrac{7}{6}} \right).\dfrac{6}{5}\\ = \dfrac{3}{2} + \left( {\dfrac{{ – 7}}{5}} \right)\\ =\dfrac{15}{{10}}+\dfrac{-14}{{10}}= \dfrac{1}{{10}}\end{array}\)