1. Quy tắc dấu ngoặc

Hoạt động khám phá 1. Tính rồi so sánh kết quả của:

a) \(\frac{3}{4} + \left ( \frac{1}{2} -\frac{1}{3}\right )\) và \(\frac{3}{4} + \frac{1}{2} - \frac{1}{3}\)

b) \(\frac{2}{3} - (\frac{1}{2} + \frac{1}{3})\) và \(\frac{2}{3} - \frac{1}{2} - \frac{1}{3}\)

Lời giải tham khảo:

a)  Ta có:

• \(\frac{3}{4} + \left ( \frac{1}{2} -\frac{1}{3}\right ) = \frac{3}{4} + \frac{1}{6} = \frac{9}{12} + \frac{2}{12} = \frac{11}{12}\).
• \(\frac{3}{4} + \frac{1}{2} - \frac{1}{3} = \frac{9}{12} + \frac{6}{12} - \frac{4}{12} = \frac{11}{12}\)

⇒ \(\frac{3}{4} + \left ( \frac{1}{2} -\frac{1}{3}\right )  = \frac{3}{4} + \frac{1}{2} - \frac{1}{3}\)

b) Ta có: 

• \(\frac{2}{3} - (\frac{1}{2} + \frac{1}{3}) = \frac{2}{3} - \frac{5}{6} = \frac{4}{6} - \frac{5}{6} = -\frac{1}{6}\)
• \(\frac{2}{3} - \frac{1}{2} - \frac{1}{3} = \frac{1}{6} - \frac{2}{6} = -\frac{1}{6}\)

⇒ \(\frac{2}{3} - (\frac{1}{2} + \frac{1}{3})\) = \(\frac{2}{3} - \frac{1}{2} - \frac{1}{3}\)

Thực hành 1: Cho biểu thức: Hãy tính giá trị của A bằng cách bỏ dấu ngoặc rồi nhóm các số hạng thích hợp.

A =\(7-\frac{2}{5}+\frac{1}{3} - 6 + \frac{4}{3} -\frac{6}{5} - 2 +\frac{8}{5} -\frac{5}{3}\)

Lời giải tham khảo:

Ta có: \(7-\frac{2}{5}+\frac{1}{3} - 6 + \frac{4}{3} -\frac{6}{5} - 2 +\frac{8}{5} -\frac{5}{3}\)

= \((7-6-2) + \left ( -\frac{2}{5} -\frac{6}{5} + \frac{8}{5}\right ) + \left ( \frac{1}{3} +\frac{4}{3} -\frac{5}{3}\right )\)

= -1 + 0+ 0 

= -1

2. Quy tắc chuyển vế

Hoạt động khám phá 2. Thực hiện bài toán tìm x, biết \(x-\frac{2}{5}=\frac{1}{2}\) theo hướng dẫn sau:

• Cộng hai vế với \(\frac{2}{5}\)
• Rút gọn hai vế
• Ghi kết quả

Lời giải tham khảo:

Ta có: \(x-\frac{2}{5}=\frac{1}{2}\)

\(Leftrightarrow x=\frac{1}{2} + \frac{2}{5}\)

\(Leftrightarrow x= \frac{9}{10}\)

Thực hành 2. Tìm x, biết:

a) \(x + \frac{1}{2}=-\frac{1}{3}\)

b) \(-\frac{2}{7} + x =-\frac{1}{4}\)

Lời giải tham khảo:

a) \(x + \frac{1}{2}=-\frac{1}{3}\)

\( \Leftrightarrow x = -\frac{1}{3} - \frac{1}{2}\)

\( \Leftrightarrow x = -\frac{5}{6}\)

b) \(-\frac{2}{7} + x =-\frac{1}{4}\)

\( \Leftrightarrow x =-\frac{1}{4} + \frac{2}{7}\)

\( \Leftrightarrow x = \frac{1}{28}\)

3. Thứ tự thực hiện các phép tính

Thực hành 3. Tính

a) \(1\frac{1}{2} + \frac{1}{5}.\left [\left ( -2\frac{5}{6}\right ) +\frac{1}{3}  \right ]\)

b) \(\frac{1}{3} . \left (\frac{2}{5} - \frac{1}{2} \right) : \left ( \frac{1}{6} - \frac{1}{5} \right )^{2}\)

Lời giải tham khảo:

a) \(1\frac{1}{2} + \frac{1}{5}.\left [\left ( -2\frac{5}{6}\right ) +\frac{1}{3}  \right ]\)

= \(\frac{3}{2} + \frac{1}{5}.\left [ -\frac{17}{6}+\frac{1}{3}  \right ]\)

= \(\frac{3}{2} + \frac{1}{5}.\left [ -\frac{17}{6}+\frac{2}{6}  \right ]\)

= \(\frac{3}{2} + \frac{1}{5}.-\frac{5}{2}\)

= \(\frac{3}{2} - \frac{1}{2}\)

= 1

b) \(\frac{1}{3} . \left (\frac{2}{5} - \frac{1}{2} \right) : \left ( \frac{1}{6} - \frac{1}{5} \right )^{2}\)

= \(\frac{1}{3} .\left (\frac{4}{10} - \frac{5}{10} \right) : \left ( \frac{5}{30} - \frac{6}{30} \right )^{2}\)

= \(\frac{1}{3} . \frac{-1}{10}  : \left (- \frac{1}{30} \right )^{2}\)

= \(-\frac{1}{30} : \frac{1}{900}\)

= -30

4. Bài tập

Câu 1 (Trang 24 SGK tập 1 Chân Trời Sáng Tạo). Bỏ dấu ngoặc rồi tính

a) \(\left ( \frac{-3}{7}\right ) + \left ( \frac{5}{6} -\frac{4}{7} \right )\)

b) \(\frac{3}{5} - \left ( \frac{2}{3} +\frac{1}{5} \right )\)

c) \(\left [ \left ( \frac{-1}{3}\right ) + 1 \right ] - \left ( \frac{2}{3} -\frac{1}{5}\right )\)

d) \(1\frac{1}{3} + \left (\frac{2}{3} -\frac{3}{4}  \right ) - \left ( 0,8 + 1\frac{1}{5} \right )\)

Lời giải tham khảo:

a) \(\left ( \frac{-3}{7}\right ) + \left ( \frac{5}{6} -\frac{4}{7} \right )\)

= \(\left ( \frac{-3}{7}\right ) + \left ( \frac{35}{42} -\frac{24}{42} \right )\)

= \(\left ( \frac{-3}{7}\right ) + \frac{11}{42}\)

= \(\left ( \frac{-18}{42}\right ) + \frac{11}{42}\)

= \(\frac{-1}{6}\)

b) \(\frac{3}{5} - \left ( \frac{2}{3} +\frac{1}{5} \right )\)

= \(\frac{3}{5} - \left ( \frac{10}{15} +\frac{3}{15} \right )\)

= \(\frac{3}{5} - \frac{13}{15}\)

= \(\frac{9}{15} - \frac{13}{15}\)

= \(\frac{-4}{15}\)

c) \(\left [ \left ( \frac{-1}{3}\right ) + 1 \right ] - \left ( \frac{2}{3} -\frac{1}{5}\right )\)

= \(\left [ \left ( \frac{-1}{3}\right ) + \frac{3}{3} \right ] - \left ( \frac{10}{15} -\frac{3}{15}\right )\)

= \(\frac{2}{3} - \frac{7}{15}\)

= \(\frac{10}{15} - \frac{7}{15}\)

= \(\frac{1}{5}\)

d) \(1\frac{1}{3} + \left (\frac{2}{3} -\frac{3}{4}  \right ) - \left ( 0,8 + 1\frac{1}{5} \right )\)

= \(\frac{4}{3} + \left (\frac{8}{12} -\frac{9}{12}  \right ) - \left ( \frac{4}{5} + \frac{6}{5} \right )\)

= \(\frac{4}{3} -\frac{1}{12}  - 2\)

= \(\frac{16}{12} -\frac{1}{12}  - \frac{24}{12}\)

= \(-\frac{9}{12}\)

= \(-\frac{3}{4}\)

Bài 2 (Trang 25 SGK tập 1 Chân Trời Sáng Tạo): 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 tham khảo:

a) \(\left( {\frac{3}{4}:1\frac{1}{2}} \right) - \left( {\frac{5}{6}:\frac{1}{3}} \right)\)

\(\begin{array}{l} = \left( {\frac{3}{4}:\frac{3}{2}} \right) - \left( {\frac{5}{6}.3} \right)\\ = \left( {\frac{3}{4}.\frac{2}{3}} \right) - \frac{5}{2}\\ = \frac{1}{2} - \frac{5}{2} \end{array}\)

= -2.

b) \(\left[ {\left( {\frac{{ - 1}}{5}} \right):\frac{1}{{10}}} \right] - \frac{5}{7}.\left( {\frac{2}{3} - \frac{1}{5}} \right)\)

\(\begin{array}{l}= \left( {\frac{{ - 1}}{5}} \right).10 - \frac{5}{7}.\left( {\frac{{10}}{{15}} - \frac{3}{{15}}} \right)\\ =  - 2 - \frac{5}{7}.\frac{7}{{15}}\\ =  - 2 - \frac{1}{3}\\ = \frac{{ - 6}}{3} - \frac{1}{3}\\ = \frac{{ - 7}}{3}\end{array}\)

c) \(\left( { - 0,4} \right) + 2\frac{2}{5}.{\left[ {\left( {\frac{{ - 2}}{3}} \right) + \frac{1}{2}} \right]^2}\)

\(\begin{array}{l} = \left( { - \frac{2}{5}} \right) + \frac{{12}}{5}.{\left[ {\left( {\frac{{ - 4}}{6}} \right) + \frac{3}{6}} \right]^2}\\ = \left( { - \frac{2}{5}} \right) + \frac{{12}}{5}.{\left( {\frac{{ - 1}}{6}} \right)^2}\\ = \left( { - \frac{2}{5}} \right) + \frac{{12}}{5}.\frac{1}{{36}}\\ = \left( { - \frac{2}{5}} \right) + \frac{1}{{15}}\\ = \left( { - \frac{6}{{15}}} \right) + \frac{1}{{15}}\\ = \frac{{ - 5}}{{15}} \\ = \frac{{ - 1}}{3}\end{array}\)

d) \(\left\{ {\left[ {{{\left( {\frac{1}{{25}} - 0,6} \right)}^2}:\frac{{49}}{{125}}} \right].\frac{5}{6}} \right\} - \left[ {\left( {\frac{{ - 1}}{3}} \right) + \frac{1}{2}} \right]\)

\(\begin{array}{l}= \left\{ {\left[ {{{\left( {\frac{1}{{25}} - \frac{3}{5}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} - \left[ {\left( {\frac{{ - 2}}{6}} \right) + \frac{3}{6}} \right]\\ = \left\{ {\left[ {{{\left( {\frac{{ - 14}}{{25}}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} - \frac{1}{6}\\ = \left\{ {\frac{{49.4}}{{{{25}^2}}}.\frac{{25.5}}{{49}}.\frac{5}{6}} \right\} - \frac{1}{6}\\ = \frac{4}{6} - \frac{1}{6} \\= \frac{3}{6} \\= \frac{1}{2}\end{array}\)

Bài 3 (Trang 25 SGK tập 1 Chân Trời Sáng Tạo): Cho biểu thức: Hãy tính giá trị của A theo hai cách:

\(A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right).\)

a) Tính giá trị của từng biểu thức trong dấu ngoặc trước.

b) Bỏ dấu ngoặc rồi nhóm các số hạng thích hợp.

Lời giải tham khảo:

a) \(A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right).\)

\(\begin{array}{l}A = \left( {\frac{{30}}{{15}} + \frac{5}{{15}} - \frac{6}{{15}}} \right) - \left( {\frac{{105}}{{15}} - \frac{9}{{15}} - \frac{{20}}{{15}}} \right) - \left( {\frac{3}{{15}} + \frac{{25}}{{15}} - \frac{{60}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} - \left( {\frac{{ - 32}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} + \frac{{32}}{{15}}\\A = \frac{{ - 15}}{{15}}\end{array}\)

A = -1.

b) \(A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right).\)

\(\begin{array}{l}A = 2 + \frac{1}{3} - \frac{2}{5} - 7 + \frac{3}{5} + \frac{4}{3} - \frac{1}{5} - \frac{5}{3} + 4\\A = \left( {2 - 7 + 4} \right) + \left( {\frac{1}{3} + \frac{4}{3} - \frac{5}{3}} \right) + \left( { - \frac{2}{5} + \frac{3}{5} - \frac{1}{5}} \right)\end{array}\)

A = -1 + 0 + 0

A = -1.

Bài 4 (Trang 25 SGK tập 1 Chân Trời Sáng Tạo): 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 tham khảo:

a) \(x + \frac{3}{5} = \frac{2}{3}\)

\(\begin{array}{l} \Leftrightarrow x = \frac{2}{3} - \frac{3}{5}\\ \Leftrightarrow x = \frac{{10}}{{15}} - \frac{9}{{15}}\\ \Leftrightarrow x = \frac{1}{{15}}\end{array}\)

b) \(\frac{3}{7} - x = \frac{2}{5}\)

\(\begin{array}{l} \Leftrightarrow x = \frac{3}{7} - \frac{2}{5}\\ \Leftrightarrow x = \frac{{15}}{{35}} - \frac{{14}}{{35}}\\ \Leftrightarrow x = \frac{1}{{35}}\end{array}\)

c) \(\frac{4}{9} - \frac{2}{3}x = \frac{1}{3}\)

\(\begin{array}{l} \Leftrightarrow \frac{2}{3}x = \frac{4}{9} - \frac{1}{3}\\ \Leftrightarrow \frac{2}{3}x = \frac{4}{9} - \frac{3}{9}\\ \Leftrightarrow \frac{2}{3}x = \frac{1}{9}\\ \Leftrightarrow x = \frac{1}{9}:\frac{2}{3}\\ \Leftrightarrow x = \frac{1}{9}.\frac{3}{2}\\ \Leftrightarrow x = \frac{1}{6}\end{array}\)

d) \(\frac{3}{{10}}x - 1\frac{1}{2} = \left( {\frac{{ - 2}}{7}} \right):\frac{5}{{14}}\)

\(\begin{array}{l} \Leftrightarrow \frac{3}{{10}}x - \frac{3}{2} = \left( {\frac{{ - 2}}{7}} \right).\frac{{14}}{5}\\ \Leftrightarrow \frac{3}{{10}}x - \frac{3}{2} = \frac{{ - 4}}{5}\\ \Leftrightarrow \frac{3}{{10}}x = \frac{{ - 4}}{5} + \frac{3}{2}\\ \Leftrightarrow \frac{3}{{10}}x = \frac{{ - 8}}{{10}} + \frac{{15}}{{10}}\\ \Leftrightarrow \frac{3}{{10}}x = \frac{7}{{10}}\\ \Leftrightarrow x = \frac{7}{{10}}:\frac{3}{{10}}\\ \Leftrightarrow x = \frac{7}{3}\end{array}\)

Bài 5 (Trang 25 SGK tập 1 Chân Trời Sáng Tạo): 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 tham khảo:

a) \(\frac{2}{9}:x + \frac{5}{6} = 0,5\)

\(\begin{array}{l}\Leftrightarrow \frac{2}{9}:x = \frac{1}{2} - \frac{5}{6}\\ \Leftrightarrow \frac{2}{9}:x = \frac{3}{6} - \frac{5}{6}\\ \Leftrightarrow \frac{2}{9}:x = \frac{{ - 2}}{6}\\ \Leftrightarrow x = \frac{2}{9}:\frac{{ - 2}}{6}\\ \Leftrightarrow x = \frac{2}{9}.\frac{{ - 6}}{2}\\ \Leftrightarrow x = \frac{{ - 2}}{3}\end{array}\)

b) \(\frac{3}{4} - \left( {x - \frac{2}{3}} \right) = 1\frac{1}{3}\)

\(\begin{array}{l} \Leftrightarrow x - \frac{2}{3} = \frac{3}{4} - 1\frac{1}{3}\\ \Leftrightarrow x - \frac{2}{3} = \frac{3}{4} - \frac{4}{3}\\ \Leftrightarrow x - \frac{2}{3} = \frac{9}{{12}} - \frac{{16}}{{12}}\\ \Leftrightarrow x - \frac{2}{3} = \frac{{ - 7}}{{12}}\\ \Leftrightarrow x = \frac{{ - 7}}{{12}} + \frac{2}{3}\\ \Leftrightarrow x = \frac{{ - 7}}{{12}} + \frac{8}{{12}}\\ \Leftrightarrow x = \frac{1}{12}\end{array}\)

c) \(1\frac{1}{4}:\left( {x - \frac{2}{3}} \right) = 0,75\)

\(\begin{array}{l} \Leftrightarrow x - \frac{2}{3} = \frac{3}{4} - 1\frac{1}{3}\\ \Leftrightarrow x - \frac{2}{3} = \frac{3}{4} - \frac{4}{3}\\ \Leftrightarrow x - \frac{2}{3} = \frac{9}{{12}} - \frac{{16}}{{12}}\\ \Leftrightarrow x - \frac{2}{3} = \frac{{ - 7}}{{12}}\\ \Leftrightarrow x = \frac{{ - 7}}{{12}} + \frac{2}{3}\\ \Leftrightarrow x = \frac{{ - 7}}{{12}} + \frac{8}{{12}}\\ \Leftrightarrow x = \frac{1}{12}\end{array}\)

d) \(\left( { - \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}\)

\(\begin{array}{l} \Leftrightarrow  - \frac{5}{6}x + \frac{5}{4} = \frac{4}{3}.\frac{3}{2}\\  \Leftrightarrow - \frac{5}{6}x + \frac{5}{4} = 2\\  \Leftrightarrow - \frac{5}{6}x = 2 - \frac{5}{4}\\  \Leftrightarrow - \frac{5}{6}x = \frac{8}{4} - \frac{5}{4}\\  \Leftrightarrow - \frac{5}{6}x = \frac{3}{4}\\ \Leftrightarrow x = \frac{3}{4}:\left( { - \frac{5}{6}} \right)\\ \Leftrightarrow x = \frac{3}{4}.\frac{{ - 6}}{5}\\ \Leftrightarrow x = \frac{{ - 9}}{{10}}\end{array}\)

Bài 6 (Trang 25 SGK tập 1 Chân Trời Sáng Tạo): 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 tham khảo:

a) \(\frac{{13}}{{23}}.\frac{7}{{11}} + \frac{{10}}{{23}}.\frac{7}{{11}}\)

\(\begin{array}{l} = \frac{7}{{11}}\left( {\frac{{13}}{{23}} + \frac{{10}}{{23}}} \right)\\ = \frac{7}{{11}}.1\\ = \frac{7}{{11}}\end{array}\)

b) \(\frac{5}{9}.\frac{{23}}{{11}} - \frac{1}{{11}}.\frac{5}{9} + \frac{5}{9}\)

\(\begin{array}{l} = \frac{5}{9}.\left( {\frac{{23}}{{11}} - \frac{1}{{11}} + 1} \right)\\ = \frac{5}{9}.\left( {2 + 1} \right)\\ = \frac{5}{9}.3 = \frac{5}{3}\end{array}\)

c) \(\left[ {\left( { - \frac{4}{9}} \right) + \frac{3}{5}} \right]:\frac{{13}}{{17}} + \left( {\frac{2}{5} - \frac{5}{9}} \right):\frac{{13}}{{17}}\)

\(\begin{array}{l}= \left( { - \frac{4}{9} + \frac{3}{5}} \right).\frac{{17}}{{13}} + \left( {\frac{2}{5} - \frac{5}{9}} \right).\frac{{17}}{{13}}\\ = \frac{{17}}{{13}}.\left( { - \frac{4}{9} + \frac{3}{5} + \frac{2}{5} - \frac{5}{9}} \right)\\ = \frac{{17}}{{13}}.\left[ {\left( { - \frac{4}{9} - \frac{5}{9}} \right) + \left( {\frac{3}{5} + \frac{2}{5}} \right)} \right]\\ = \frac{{17}}{{13}}.\left( { - 1 + 1} \right)\\ = \frac{{17}}{{13}}.0\end{array}\)

= 0

d) \(\frac{3}{{16}}:\left( {\frac{3}{{22}} - \frac{3}{{11}}} \right) + \frac{3}{{16}}:\left( {\frac{1}{{10}} - \frac{2}{5}} \right)\)

\(\begin{array}{l} = \frac{3}{{16}}:\left( {\frac{3}{{22}} - \frac{6}{{22}}} \right) + \frac{3}{{16}}:\left( {\frac{1}{{10}} - \frac{4}{{10}}} \right)\\ = \frac{3}{{16}}:\frac{{ - 3}}{{22}} + \frac{3}{{16}}:\frac{{ - 3}}{{10}}\\ = \frac{3}{{16}}.\frac{{ - 22}}{3} + \frac{3}{{16}}.\frac{{ - 10}}{3}\\ = \frac{3}{{16}}.\left( {\frac{{ - 22}}{3} + \frac{{ - 10}}{3}} \right)\\ = \frac{3}{{16}}.\frac{{ - 32}}{3}\end{array}\)

= -2

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