Giải bài tập trang 24, 25 Bài 4 Quy tắc dấu ngoặc và quy tắc chuyển vế sgk toán 7 tập 1 chân trời sáng tạo. Bài 1 Bỏ dấu ngoặc rồi tính.
Bài 1 trang 24 sách giáo khoa Toán 7 Chân trời sáng tạo tập 1
Bỏ dấu ngoặc rồi tính:
a)\(\left( {\frac{{ – 3}}{7}} \right) + \left( {\frac{5}{6} – \frac{4}{7}} \right);\)
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b)\(\frac{3}{5} – \left( {\frac{2}{3} + \frac{1}{5}} \right);\)
c)\(\left[ {\left( {\frac{{ – 1}}{3} + 1} \right) – \left( {\frac{2}{3} – \frac{1}{5}} \right)} \right];\)
d)\(1\frac{1}{3} + \left( {\frac{2}{3} – \frac{3}{4}} \right) – \left( {0,8 + 1\frac{1}{5}} \right)\).
Lời giải:
a)
\(\begin{array}{l}\left( {\frac{{ – 3}}{7}} \right) + \left( {\frac{5}{6} – \frac{4}{7}} \right)\\ = \left( {\frac{{ – 3}}{7}} \right) + \frac{5}{6} – \frac{4}{7}\\ = \left[ {\left( {\frac{{ – 3}}{7}} \right) – \frac{4}{7}} \right] + \frac{5}{6}\\ =\frac{-7}{7}+\frac{5}{6}\\= – 1 + \frac{5}{6}\\ = \frac{{ – 1}}{6}\end{array}\)
b)
\(\begin{array}{l}\frac{3}{5} – \left( {\frac{2}{3} + \frac{1}{5}} \right)\\ = \frac{3}{5} – \frac{2}{3} – \frac{1}{5}\\ = (\frac{3}{5} – \frac{1}{5}) – \frac{2}{3}\\ = \frac{2}{5} – \frac{2}{3}\\ = \frac{6}{{15}} – \frac{{10}}{{15}}\\ = \frac{{ – 4}}{{15}}\end{array}\)
c)
\(\begin{array}{l}\left[ {\left( {\frac{{ – 1}}{3}} \right) + 1} \right] – \left( {\frac{2}{3} – \frac{1}{5}} \right)\\ = \left( {\frac{{ – 1}}{3}} \right) + 1 – \frac{2}{3} + \frac{1}{5}\\ = \left( {\frac{{ – 1}}{3} – \frac{2}{3}} \right) + 1 + \frac{1}{5}\\ = \frac{-3}{3}+1+\frac{1}{5}\\= – 1 + 1 + \frac{1}{5}\\ = \frac{1}{5}\end{array}\)
d)
\(\begin{array}{l}1\frac{1}{3} + \left( {\frac{2}{3} – \frac{3}{4}} \right) – \left( {0,8 + 1\frac{1}{5}} \right)\\ = 1 + \frac{1}{3} + \frac{2}{3} – \frac{3}{4} – \left( {\frac{4}{5} + 1 + \frac{1}{5}} \right)\\=1+\frac{3}{3}-\frac{3}{4}-(\frac{5}{5}+1)\\ = 1 + 1 – \frac{3}{4} – (1+1)\\ = – \frac{3}{4}\end{array}\).
Bài 2 trang 25 sách giáo khoa Toán 7 Chân trời sáng tạo tập 1
Tính:
a) \(\left( {\frac{3}{4}:1\frac{1}{2}} \right) – \left( {\frac{5}{6}:\frac{1}{3}} \right)\)
b) \(\left[ {\left( {\frac{{ – 1}}{5}} \right):\frac{1}{{10}}} \right] – \frac{5}{7}.\left( {\frac{2}{3} – \frac{1}{5}} \right)\)
c) \(\left( { – 0,4} \right) + 2\frac{2}{5}.{\left[ {\left( {\frac{{ – 2}}{3}} \right) + \frac{1}{2}} \right]^2}\)
d)\(\left\{ {\left[ {{{\left( {\frac{1}{{25}} – 0,6} \right)}^2}:\frac{{49}}{{125}}} \right].\frac{5}{6}} \right\} – \left[ {\left( {\frac{{ – 1}}{3}} \right) + \frac{1}{2}} \right]\)
Lời giải:
a)
\(\begin{array}{l}\left( {\frac{3}{4}:1\frac{1}{2}} \right) – \left( {\frac{5}{6}:\frac{1}{3}} \right)\\ = \left( {\frac{3}{4}:\frac{3}{2}} \right) – \left( {\frac{5}{6}.3} \right)\\ = \left( {\frac{3}{4}.\frac{2}{3}} \right) – \frac{5}{2}\\ = \frac{1}{2} – \frac{5}{2}\\ = \frac{-4}{2}\\= – 2.\end{array}\)
b)
\(\begin{array}{l}\left[ {\left( {\frac{{ – 1}}{5}} \right):\frac{1}{{10}}} \right] – \frac{5}{7}.\left( {\frac{2}{3} – \frac{1}{5}} \right)\\ = \left( {\frac{{ – 1}}{5}} \right).10 – \frac{5}{7}.\left( {\frac{{10}}{{15}} – \frac{3}{{15}}} \right)\\ = – 2 – \frac{5}{7}.\frac{7}{{15}}\\ = – 2 – \frac{1}{3}\\ = \frac{{ – 6}}{3} – \frac{1}{3}\\ = \frac{{ – 7}}{3}\end{array}\)
c)
\(\begin{array}{l}\left( { – 0,4} \right) + 2\frac{2}{5}.{\left[ {\left( {\frac{{ – 2}}{3}} \right) + \frac{1}{2}} \right]^2}\\ = \left( { – \frac{2}{5}} \right) + \frac{{12}}{5}.{\left[ {\left( {\frac{{ – 4}}{6}} \right) + \frac{3}{6}} \right]^2}\\ = \left( { – \frac{2}{5}} \right) + \frac{{12}}{5}.{\left( {\frac{{ – 1}}{6}} \right)^2}\\ = \left( { – \frac{2}{5}} \right) + \frac{{12}}{5}.\frac{1}{{36}}\\ = \left( { – \frac{2}{5}} \right) + \frac{1}{{15}}\\ = \left( { – \frac{6}{{15}}} \right) + \frac{1}{{15}}\\ = \frac{{ – 5}}{{15}}\\ = \frac{{ – 1}}{3}\end{array}\)
d)
\(\begin{array}{l}\left\{ {\left[ {{{\left( {\frac{1}{{25}} – 0,6} \right)}^2}:\frac{{49}}{{125}}} \right].\frac{5}{6}} \right\} – \left[ {\left( {\frac{{ – 1}}{3}} \right) + \frac{1}{2}} \right]\\ = \left\{ {\left[ {{{\left( {\frac{1}{{25}} – \frac{3}{5}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} – \left[ {\left( {\frac{{ – 2}}{6}} \right) + \frac{3}{6}} \right]\\ = \left\{ {\left[ {{{\left( {\frac{{ 1}}{{25}}-\frac{15}{25}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} – \frac{1}{6}\\ = \left\{ {\left[ {{{\left( {\frac{{ – 14}}{{25}}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} – \frac{1}{6}\\ = \left\{ {\frac{{196}}{{{{25}^2}}}.\frac{{25.5}}{{49}}.\frac{5}{6}} \right\} – \frac{1}{6}\\ = \left( {\frac{{4.49.25.5.5}}{{{{25}^2}.49.6}}} \right) – \frac{1}{6}\\ = \frac{4}{6} – \frac{1}{6}\\ = \frac{3}{6}\\ = \frac{1}{2}\end{array}\)
Bài 3 trang 25 sách giáo khoa Toán 7 Chân trời sáng tạo tập 1
Cho biểu thức: \(A = \left( {2 + \frac{1}{3} – \frac{2}{5}} \right) – \left( {7 – \frac{3}{5} – \frac{4}{3}} \right) – \left( {\frac{1}{5} + \frac{5}{3} – 4} \right).\)
Hãy tính giá trị của A theo hai cách:
a) Tính giá trị của từng biểu thức trong dấu ngoặc trước.
b) Bỏ dấu ngoặc rồi nhóm các số hạng thích hợp.
Lời giải:
a)
\(\begin{array}{l}A = \left( {2 + \frac{1}{3} – \frac{2}{5}} \right) – \left( {7 – \frac{3}{5} – \frac{4}{3}} \right) – \left( {\frac{1}{5} + \frac{5}{3} – 4} \right).\\A = \left( {\frac{{30}}{{15}} + \frac{5}{{15}} – \frac{6}{{15}}} \right) – \left( {\frac{{105}}{{15}} – \frac{9}{{15}} – \frac{{20}}{{15}}} \right) – \left( {\frac{3}{{15}} + \frac{{25}}{{15}} – \frac{{60}}{{15}}} \right)\\A = \frac{{29}}{{15}} – \frac{{76}}{{15}} – \left( {\frac{{ – 32}}{{15}}} \right)\\A = \frac{{29}}{{15}} – \frac{{76}}{{15}} + \frac{{32}}{{15}}\\A = \frac{{ – 15}}{{15}}\\A = – 1\end{array}\)
b)
\(\begin{array}{l}A = \left( {2 + \frac{1}{3} – \frac{2}{5}} \right) – \left( {7 – \frac{3}{5} – \frac{4}{3}} \right) – \left( {\frac{1}{5} + \frac{5}{3} – 4} \right)\\A = 2 + \frac{1}{3} – \frac{2}{5} – 7 + \frac{3}{5} + \frac{4}{3} – \frac{1}{5} – \frac{5}{3} + 4\\A = \left( {2 – 7 + 4} \right) + \left( {\frac{1}{3} + \frac{4}{3} – \frac{5}{3}} \right) + \left( { – \frac{2}{5} + \frac{3}{5} – \frac{1}{5}} \right)\\A = – 1 + 0 + 0 = – 1\end{array}\)
Bài 4 trang 25 sách giáo khoa Toán 7 Chân trời sáng tạo tập 1
Tìm x, biết:
a)\(x + \frac{3}{5} = \frac{2}{3};\)
b)\(\frac{3}{7} – x = \frac{2}{5};\)
c)\(\frac{4}{9} – \frac{2}{3}x = \frac{1}{3};\)
d)\(\frac{3}{{10}}x – 1\frac{1}{2} = \left( {\frac{{ – 2}}{7}} \right):\frac{5}{{14}}\)
Lời giải:
a)
\(\begin{array}{l}x + \frac{3}{5} = \frac{2}{3}\\x = \frac{2}{3} – \frac{3}{5}\\x = \frac{{10}}{{15}} – \frac{9}{{15}}\\x = \frac{1}{{15}}\end{array}\)
Vậy \(x = \frac{1}{{15}}\).
b)
\(\begin{array}{l}\frac{3}{7} – x = \frac{2}{5}\\x = \frac{3}{7} – \frac{2}{5}\\x = \frac{{15}}{{35}} – \frac{{14}}{{35}}\\x = \frac{1}{{35}}\end{array}\)
Vậy \(x = \frac{1}{{35}}\).
c)
\(\begin{array}{l}\frac{4}{9} – \frac{2}{3}x = \frac{1}{3}\\\frac{2}{3}x = \frac{4}{9} – \frac{1}{3}\\\frac{2}{3}x = \frac{4}{9} – \frac{3}{9}\\\frac{2}{3}x = \frac{1}{9}\\x = \frac{1}{9}:\frac{2}{3}\\x = \frac{1}{9}.\frac{3}{2}\\x = \frac{1}{6}\end{array}\)
Vậy \(x = \frac{1}{6}\).
d)
\(\begin{array}{l}\frac{3}{{10}}x – 1\frac{1}{2} = \left( {\frac{{ – 2}}{7}} \right):\frac{5}{{14}}\\\frac{3}{{10}}x – \frac{3}{2} = \left( {\frac{{ – 2}}{7}} \right).\frac{{14}}{5}\\\frac{3}{{10}}x – \frac{3}{2} = \frac{{ – 4}}{5}\\\frac{3}{{10}}x = \frac{{ – 4}}{5} + \frac{3}{2}\\\frac{3}{{10}}x = \frac{{ – 8}}{{10}} + \frac{{15}}{{10}}\\\frac{3}{{10}}x = \frac{7}{{10}}\\x = \frac{7}{{10}}:\frac{3}{{10}}\\x = \frac{7}{3}\end{array}\)
Vậy \(x = \frac{7}{3}\).
Bài 5 trang 25 sách giáo khoa Toán 7 Chân trời sáng tạo tập 1
Tìm x, biết:
a)\(\frac{2}{9}:x + \frac{5}{6} = 0,5;\)
b)\(\frac{3}{4} – \left( {x – \frac{2}{3}} \right) = 1\frac{1}{3};\)
c)\(1\frac{1}{4}:\left( {x – \frac{2}{3}} \right) = 0,75;\)
d)\(\left( { – \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}\).
Lời giải:
a)
\(\begin{array}{l}\frac{2}{9}:x + \frac{5}{6} = 0,5\\\frac{2}{9}:x = \frac{1}{2} – \frac{5}{6}\\\frac{2}{9}:x = \frac{3}{6} – \frac{5}{6}\\\frac{2}{9}:x = \frac{{ – 2}}{6}\\x = \frac{2}{9}:\frac{{ – 2}}{6}\\x = \frac{2}{9}.\frac{{ – 6}}{2}\\x = \frac{{ – 2}}{3}\end{array}\)
Vậy \(x = \frac{{ – 2}}{3}\).
b)
\(\begin{array}{l}\frac{3}{4} – \left( {x – \frac{2}{3}} \right) = 1\frac{1}{3}\\x – \frac{2}{3} = \frac{3}{4} – 1\frac{1}{3}\\x – \frac{2}{3} = \frac{3}{4} – \frac{4}{3}\\x – \frac{2}{3} = \frac{9}{{12}} – \frac{{16}}{{12}}\\x – \frac{2}{3} = \frac{{ – 7}}{{12}}\\x = \frac{{ – 7}}{{12}} + \frac{2}{3}\\x = \frac{{ – 7}}{{12}} + \frac{8}{{12}}\\x = \frac{1}{12}\end{array}\)
Vậy\(x = \frac{1}{12}\).
c)
\(\begin{array}{l}1\frac{1}{4}:\left( {x – \frac{2}{3}} \right) = 0,75\\\frac{5}{4}:\left( {x – \frac{2}{3}} \right) = \frac{3}{4}\\x – \frac{2}{3} = \frac{5}{4}:\frac{3}{4}\\x – \frac{2}{3} = \frac{5}{4}.\frac{4}{3}\\x – \frac{2}{3} = \frac{5}{3}\\x = \frac{5}{3} + \frac{2}{3}\\x = \frac{7}{3}\end{array}\)
Vậy \(x = \frac{7}{3}\).
d)
\(\begin{array}{l}\left( { – \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}\\ – \frac{5}{6}x + \frac{5}{4} = \frac{4}{3}.\frac{3}{2}\\ – \frac{5}{6}x + \frac{5}{4} = 2\\ – \frac{5}{6}x = 2 – \frac{5}{4}\\ – \frac{5}{6}x = \frac{8}{4} – \frac{5}{4}\\ – \frac{5}{6}x = \frac{3}{4}\\x = \frac{3}{4}:\left( { – \frac{5}{6}} \right)\\x = \frac{3}{4}.\frac{{ – 6}}{5}\\x = \frac{{ – 9}}{{10}}\end{array}\)
Vậy \(x = \frac{{ – 9}}{{10}}\).
Bài 6 trang 25 sách giáo khoa Toán 7 Chân trời sáng tạo tập 1
Tính nhanh:
a)\(\frac{{13}}{{23}}.\frac{7}{{11}} + \frac{{10}}{{23}}.\frac{7}{{11}};\)
b) \(\frac{5}{9}.\frac{{23}}{{11}} – \frac{1}{{11}}.\frac{5}{9} + \frac{5}{9}\)
c)\(\left[ {\left( { – \frac{4}{9}} \right) + \frac{3}{5}} \right]:\frac{{13}}{{17}} + \left( {\frac{2}{5} – \frac{5}{9}} \right):\frac{{13}}{{17}};\)
d) \(\frac{3}{{16}}:\left( {\frac{3}{{22}} – \frac{3}{{11}}} \right) + \frac{3}{{16}}:\left( {\frac{1}{{10}} – \frac{2}{5}} \right)\)
Lời giải:
a)
\(\begin{array}{l}\frac{{13}}{{23}}.\frac{7}{{11}} + \frac{{10}}{{23}}.\frac{7}{{11}}\\ = \frac{7}{{11}}\left( {\frac{{13}}{{23}} + \frac{{10}}{{23}}} \right)\\ = \frac{7}{{11}}.1\\ = \frac{7}{{11}}\end{array}\)
b)
\(\begin{array}{l}\frac{5}{9}.\frac{{23}}{{11}} – \frac{1}{{11}}.\frac{5}{9} + \frac{5}{9}\\ = \frac{5}{9}.\left( {\frac{{23}}{{11}} – \frac{1}{{11}} + 1} \right)\\ = \frac{5}{9}.\left( {2 + 1} \right)\\ = \frac{5}{9}.3 = \frac{5}{3}\end{array}\)
c)
\(\begin{array}{l}\left[ {\left( { – \frac{4}{9} + \frac{3}{5}} \right):\frac{{13}}{{17}}} \right] + \left( {\frac{2}{5} – \frac{5}{9}} \right):\frac{{13}}{{17}}\\ = \left( { – \frac{4}{9} + \frac{3}{5}} \right).\frac{{17}}{{13}} + \left( {\frac{2}{5} – \frac{5}{9}} \right).\frac{{17}}{{13}}\\ = \frac{{17}}{{13}}.\left( { – \frac{4}{9} + \frac{3}{5} + \frac{2}{5} – \frac{5}{9}} \right)\\ = \frac{{17}}{{13}}.\left[ {\left( { – \frac{4}{9} – \frac{5}{9}} \right) + \left( {\frac{3}{5} + \frac{2}{5}} \right)} \right]\\ = \frac{{17}}{{13}}.\left( { – 1 + 1} \right)\\ = \frac{{17}}{{13}}.0 = 0\end{array}\)
d)
\(\begin{array}{l}\frac{3}{{16}}:\left( {\frac{3}{{22}} – \frac{3}{{11}}} \right) + \frac{3}{{16}}:\left( {\frac{1}{{10}} – \frac{2}{5}} \right)\\ = \frac{3}{{16}}:\left( {\frac{3}{{22}} – \frac{6}{{22}}} \right) + \frac{3}{{16}}:\left( {\frac{1}{{10}} – \frac{4}{{10}}} \right)\\ = \frac{3}{{16}}:\frac{{ – 3}}{{22}} + \frac{3}{{16}}:\frac{{ – 3}}{{10}}\\ = \frac{3}{{16}}.\frac{{ – 22}}{3} + \frac{3}{{16}}.\frac{{ – 10}}{3}\\ = \frac{3}{{16}}.\left( {\frac{{ – 22}}{3} + \frac{{ – 10}}{3}} \right)\\ = \frac{3}{{16}}.\frac{{ – 32}}{3}\\ = – 2\end{array}\)
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