Chương 1 – Bài 4: Quy tắc dấu ngoặc và quy tắc chuyển vế

Chương \(1\) – Bài \(4:\) Quy tắc dấu ngoặc và quy tắc chuyển vế trang \(17\) vở bài tập toán lớp \(7\) tập \(1\) NXB Chân Trời Sáng Tạo.

\(1.\) Bỏ dấu ngoặc rồi tính.

a) \(\left(\displaystyle\frac{-3}{8}\right)+\left(\displaystyle\frac{7}{9}-\displaystyle\frac{5}{8}\right); \hspace{3,5cm} b) \displaystyle\frac{4}{9}-\left(\displaystyle\frac{3}{7}+\displaystyle\frac{2}{9}\right);\)

c) \(\left[\left(\displaystyle\frac{-2}{5}\right)+\displaystyle\frac{1}{3}\right]-\left(\displaystyle\frac{3}{5}-\displaystyle\frac{1}{4}\right); \quad \qquad d) \left(1\displaystyle\frac{1}{2}-\displaystyle\frac{3}{4}\right)-\left(0,25+\displaystyle\frac{1}{2}\right).\)

Giải

a) \(\left(\displaystyle\frac{-3}{8}\right)+\left(\displaystyle\frac{7}{9}-\displaystyle\frac{5}{8}\right)=\left(\displaystyle\frac{-3}{8}\right)+\displaystyle\frac{7}{9}-\displaystyle\frac{5}{8}\)

\(=\displaystyle\frac{-3}{8}+\left(\displaystyle\frac{-5}{8}\right)+\displaystyle\frac{7}{9}=-1+\displaystyle\frac{7}{9}=\displaystyle\frac{-9}{9}+\displaystyle\frac{7}{9}=\displaystyle\frac{-2}{9};\)

b) \(\displaystyle\frac{4}{9}-\left(\displaystyle\frac{3}{7}+\displaystyle\frac{2}{9}\right)=\displaystyle\frac{4}{9}-\displaystyle\frac{3}{7}-\displaystyle\frac{2}{9}\)

\(=\displaystyle\frac{4}{9}-\displaystyle\frac{2}{9}-\displaystyle\frac{3}{7}=\displaystyle\frac{2}{9}-\displaystyle\frac{3}{7}=\displaystyle\frac{14}{63}-\displaystyle\frac{27}{63}=\displaystyle\frac{-13}{63};\)

c) \(\left[\left(\displaystyle\frac{-2}{5}\right)+\displaystyle\frac{1}{3}\right]-\left(\displaystyle\frac{3}{5}-\displaystyle\frac{1}{4}\right)=\left(\displaystyle\frac{-2}{5}\right)+\displaystyle\frac{1}{3}-\displaystyle\frac{3}{5}+\displaystyle\frac{1}{4}\)

\(=\left(\displaystyle\frac{-2}{5}\right)-\displaystyle\frac{3}{5}+\displaystyle\frac{1}{3}+\displaystyle\frac{1}{4}=(-1)+\displaystyle\frac{1}{3}+\displaystyle\frac{1}{4}=\left(\displaystyle\frac{-12}{12}\right)+\displaystyle\frac{4}{12}+\displaystyle\frac{3}{12}=\displaystyle\frac{-5}{12};\)

d) \(\left(1\displaystyle\frac{1}{2}-\displaystyle\frac{3}{4}\right)-\left(0,25+\displaystyle\frac{1}{2}\right)=\displaystyle\frac{3}{2}-\displaystyle\frac{3}{4}-\displaystyle\frac{1}{4}-\displaystyle\frac{1}{2}\)

\(=\displaystyle\frac{3}{2}-\displaystyle\frac{1}{2}-\displaystyle\frac{3}{4}-\displaystyle\frac{1}{4}=0\)

\(\)

\(2.\) Tính:

a) \((-0,5)-\left(-1+\displaystyle\frac{2}{3}\right):1,5+\left(\displaystyle\frac{-1}{4}\right);\)

b) \(\left[\left(\displaystyle\frac{-7}{8}\right):\displaystyle\frac{21}{16}\right]-\displaystyle\frac{-5}{3}.\left(\displaystyle\frac{1}{3}-\displaystyle\frac{7}{10}\right);\)

c) \(\left[\left(\displaystyle\frac{-2}{3}\right)+\displaystyle\frac{3}{4}\right]^2.\displaystyle\frac{12}{5}-\displaystyle\frac{1}{5};\)

d) \(\left(\displaystyle\frac{1}{25}-0,4\right)^2:\displaystyle\frac{9}{125}-\left[\left(1\displaystyle\frac{1}{3}-\displaystyle\frac{2}{5}\right).\displaystyle\frac{3}{7}\right];\)

e) \(\left\{3\displaystyle\frac{17}{18}.\left[\displaystyle\frac{5}{2}-\left(\displaystyle\frac{1}{3}+\displaystyle\frac{2}{9}\right)\right]\right\}:\left[\left(\displaystyle\frac{-1}{2}\right)+0,25\right]^2.\)

Giải

a) \((-0,5)-\left(-1+\displaystyle\frac{2}{3}\right):1,5+\left(\displaystyle\frac{-1}{4}\right)=\left(\displaystyle\frac{-1}{2}\right)-\left(\displaystyle\frac{-3}{3}+\displaystyle\frac{2}{3}\right):\displaystyle\frac{3}{2}+\left(\displaystyle\frac{-1}{4}\right)\)

\(=\left(\displaystyle\frac{-1}{2}\right)-\left(\displaystyle\frac{-1}{3}\right).\displaystyle\frac{2}{3}+\left(\displaystyle\frac{-1}{4}\right)=\left(\displaystyle\frac{-1}{2}\right)+\displaystyle\frac{2}{9}+\left(\displaystyle\frac{-1}{4}\right)\)

\(=\left(\displaystyle\frac{-18}{36}\right)+\displaystyle\frac{8}{36}+\left(\displaystyle\frac{-9}{36}\right)=\displaystyle\frac{-19}{36};\)

b) \(\left[\left(\displaystyle\frac{-7}{8}\right):\displaystyle\frac{21}{16}\right]-\displaystyle\frac{-5}{3}.\left(\displaystyle\frac{1}{3}-\displaystyle\frac{7}{10}\right)=\left[\left(\displaystyle\frac{-7}{8}\right).\displaystyle\frac{16}{21}\right]-\displaystyle\frac{-5}{3}.\left(\displaystyle\frac{10}{30}-\displaystyle\frac{21}{30}\right)\)

\(=\displaystyle\frac{-2}{3}-\displaystyle\frac{5}{3}.\left(\displaystyle\frac{-11}{30}\right)=\displaystyle\frac{-2}{3}+\displaystyle\frac{11}{18}=\displaystyle\frac{-1}{18};\)

c) \(\left[\left(\displaystyle\frac{-2}{3}\right)+\displaystyle\frac{3}{4}\right]^2.\displaystyle\frac{12}{5}-\displaystyle\frac{1}{5}=\left[\left(\displaystyle\frac{-8}{12}\right)+\displaystyle\frac{9}{12}\right]^2.\displaystyle\frac{12}{5}-\displaystyle\frac{1}{5}\)

\(=\left(\displaystyle\frac{1}{12}\right)^2.\displaystyle\frac{12}{5}-\displaystyle\frac{1}{5}=\displaystyle\frac{1}{60}-\displaystyle\frac{1}{5}=\displaystyle\frac{-11}{60};\)

d) \(\left(\displaystyle\frac{1}{25}-0,4\right)^2:\displaystyle\frac{9}{125}-\left[\left(1\displaystyle\frac{1}{3}-\displaystyle\frac{2}{5}\right).\displaystyle\frac{3}{7}\right]=\left(\displaystyle\frac{1}{25}-\displaystyle\frac{2}{5}\right)^2.\displaystyle\frac{125}{9}-\left[\left(\displaystyle\frac{4}{3}-\displaystyle\frac{2}{5}\right).\displaystyle\frac{3}{7}\right]\)

\(=\left(\displaystyle\frac{-9}{25}\right)^2.\displaystyle\frac{125}{9}-\left[\displaystyle\frac{14}{15}.\displaystyle\frac{3}{7}\right]=\displaystyle\frac{9^2}{5^4}.\displaystyle\frac{5^3}{9}-\displaystyle\frac{2}{5}=\displaystyle\frac{9}{5}-\displaystyle\frac{2}{5}=\displaystyle\frac{7}{5};\)

e) \(\left\{3\displaystyle\frac{17}{18}.\left[\displaystyle\frac{5}{2}-\left(\displaystyle\frac{1}{3}+\displaystyle\frac{2}{9}\right)\right]\right\}:\left[\left(\displaystyle\frac{-1}{2}\right)+0,25\right]^2=\left\{\displaystyle\frac{71}{18}.\left[\displaystyle\frac{5}{2}-\displaystyle\frac{5}{9}\right]\right\}:\left[\left(\displaystyle\frac{-1}{2}\right)+\displaystyle\frac{1}{4}\right]^2\)

\(=\left\{\displaystyle\frac{71}{18}.\displaystyle\frac{35}{18}\right\}:\left[\displaystyle\frac{-1}{4}\right]^2=\displaystyle\frac{2485}{324}:\displaystyle\frac{1}{16}=\displaystyle\frac{2485}{324}.16=\displaystyle\frac{9940}{81}\)

\(\)

\(3.\) Cho biểu thức:

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

Hãy tính giá trị của A theo hai cách:

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

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

Giải

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

A\(=\left(\displaystyle\frac{48}{6}-\displaystyle\frac{4}{6}+\displaystyle\frac{3}{6}\right)-\left(\displaystyle\frac{30}{6}-\displaystyle\frac{14}{6}-\displaystyle\frac{9}{6}\right)-\left(\displaystyle\frac{10}{6}+\displaystyle\frac{15}{6}+\displaystyle\frac{24}{6}\right)\)

A\(=\displaystyle\frac{47}{6}-\displaystyle\frac{7}{6}-\displaystyle\frac{49}{6}=\displaystyle\frac{-9}{6}=\displaystyle\frac{-3}{2}.\)

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

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

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

A\(=(-1)+0-\displaystyle\frac{1}{2}=\displaystyle\frac{-3}{2.}\)   

\(\)

\(4.\)Tìm x, biết:

a) \(x+\displaystyle\frac{3}{7}=\displaystyle\frac{2}{5};\quad \qquad b) \displaystyle\frac{3}{2}-x=\displaystyle\frac{4}{5};\)

c) \(\displaystyle\frac{5}{9}-\displaystyle\frac{1}{3}x=\displaystyle\frac{2}{3}; \qquad d) \displaystyle\frac{3}{5}x-1\displaystyle\frac{1}{5}=\displaystyle\frac{-3}{14}:\displaystyle\frac{5}{7}.\)

Giải

a) \(x+\displaystyle\frac{3}{7}=\displaystyle\frac{2}{5}\)

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

\(x=\displaystyle\frac{14}{35}-\displaystyle\frac{15}{35}=\displaystyle\frac{-1}{35}.\)

b) \(\displaystyle\frac{3}{2}-x=\displaystyle\frac{4}{5}\)

\(x=\displaystyle\frac{3}{2}-\displaystyle\frac{4}{5}\)

\(x=\displaystyle\frac{15}{10}-\displaystyle\frac{8}{10}=\displaystyle\frac{7}{10}.\)

c) \(\displaystyle\frac{5}{9}-\displaystyle\frac{1}{3}x=\displaystyle\frac{2}{3}\)

\(\displaystyle\frac{1}{3}x=\displaystyle\frac{5}{9}-\displaystyle\frac{2}{3}\)

\(\displaystyle\frac{1}{3}x=\displaystyle\frac{-1}{9}\)

\(x=\displaystyle\frac{-1}{9}:\displaystyle\frac{1}{3}=\displaystyle\frac{-1}{3}\)

d) \(\displaystyle\frac{3}{5}x-1\displaystyle\frac{1}{5}=\displaystyle\frac{-3}{14}:\displaystyle\frac{5}{7}\)

\(\displaystyle\frac{3}{5}x-\displaystyle\frac{6}{5}=\displaystyle\frac{-3}{10}\)

\(\displaystyle\frac{3}{5}x=\displaystyle\frac{-3}{10}+\displaystyle\frac{6}{5}\)

\(\displaystyle\frac{3}{5}x=\displaystyle\frac{9}{10}\)

\(x=\displaystyle\frac{9}{10}:\displaystyle\frac{3}{5}=\displaystyle\frac{3}{2}\)

\(\)

\(5.\) Tìm x, biết:

a) \(\displaystyle\frac{3}{4}+\displaystyle\frac{1}{9}:x=0,5; \hspace{3cm} b) \displaystyle\frac{3}{4}-\left(x-\displaystyle\frac{2}{3}\right)=1\displaystyle\frac{1}{3};\)

c) \(\left(\displaystyle\frac{5}{7}-x\right).\displaystyle\frac{11}{15}=\displaystyle\frac{-22}{45}; \quad \qquad d) \left(2,5x-\displaystyle\frac{4}{7}\right):\displaystyle\frac{8}{21}=-1,5.\)

Giải

a) \(\displaystyle\frac{3}{4}+\displaystyle\frac{1}{9}:x=0,5\)

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

\(\displaystyle\frac{1}{9}:x=\displaystyle\frac{-1}{4}\)

\(x=\displaystyle\frac{1}{9}:\left(\displaystyle\frac{-1}{4}\right)=\displaystyle\frac{-4}{9}.\)

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

\(x-\displaystyle\frac{2}{3}=\displaystyle\frac{3}{4}-\displaystyle\frac{4}{3}\)

\(x-\displaystyle\frac{2}{3}=\displaystyle\frac{-7}{12}\)

\(x=\displaystyle\frac{-7}{12}+\displaystyle\frac{2}{3}=\displaystyle\frac{1}{12}.\)

c) \(\left(\displaystyle\frac{5}{7}-x\right).\displaystyle\frac{11}{15}=\displaystyle\frac{-22}{45}\)

\(\displaystyle\frac{5}{7}-x=\displaystyle\frac{-22}{45}:\displaystyle\frac{11}{15}\)

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

\(x=\displaystyle\frac{5}{7}-\left(\displaystyle\frac{-2}{3}\right)=\displaystyle\frac{29}{21}.\)

d) \(\left(2,5x-\displaystyle\frac{4}{7}\right):\displaystyle\frac{8}{21}=-1,5\)

\(\left(\displaystyle\frac{5}{2}x-\displaystyle\frac{4}{7}\right):\displaystyle\frac{8}{21}=\displaystyle\frac{-3}{2}\)

\(\displaystyle\frac{5}{2}x-\displaystyle\frac{4}{7}=\displaystyle\frac{-3}{2}.\displaystyle\frac{8}{21}\)

\(\displaystyle\frac{5}{2}x-\displaystyle\frac{4}{7}=\displaystyle\frac{-4}{7}\)

\(\displaystyle\frac{5}{2}x=-\displaystyle\frac{4}{7}+\displaystyle\frac{-4}{7}\)

\(\displaystyle\frac{5}{2}x=0\)

\(x=0.\)

\(\)

\(6.\) Tính nhanh.

a) \(\displaystyle\frac{12}{23}.\displaystyle\frac{7}{13}+\displaystyle\frac{11}{23}.\displaystyle\frac{7}{13}; \hspace{2cm} b) \displaystyle\frac{4}{9}.\displaystyle\frac{23}{11}-\displaystyle\frac{1}{11}.\displaystyle\frac{4}{9}+\displaystyle\frac{4}{9};\)

c) \(\left[\left(-\displaystyle\frac{5}{7}\right)+\displaystyle\frac{3}{5}\right]:\displaystyle\frac{2020}{2021}+\left(\displaystyle\frac{2}{5}-\displaystyle\frac{2}{7}\right):\displaystyle\frac{2020}{2021};\)

d) \(\displaystyle\frac{3}{8}:\left(\displaystyle\frac{7}{22}-\displaystyle\frac{2}{11}\right)+\displaystyle\frac{3}{8}:\left(\displaystyle\frac{2}{5}-\displaystyle\frac{1}{10}\right).\)

Giải

a) \(\displaystyle\frac{12}{23}.\displaystyle\frac{7}{13}+\displaystyle\frac{11}{23}.\displaystyle\frac{7}{13}\)

\(=\displaystyle\frac{7}{13}.\left(\displaystyle\frac{12}{23}+\displaystyle\frac{11}{23}\right)\)

\(=\displaystyle\frac{7}{13}.1=\displaystyle\frac{7}{13}.\)

b) \(\displaystyle\frac{4}{9}.\displaystyle\frac{23}{11}-\displaystyle\frac{1}{11}.\displaystyle\frac{4}{9}+\displaystyle\frac{4}{9}\)

\(=\displaystyle\frac{4}{9}.\left(\displaystyle\frac{23}{11}-\displaystyle\frac{1}{11}+1\right)\)

\(\displaystyle\frac{4}{9}.(2+1)=\displaystyle\frac{4}{9}.3=\displaystyle\frac{4}{3}.\)

c) \(\left[\left(-\displaystyle\frac{5}{7}\right)+\displaystyle\frac{3}{5}\right]:\displaystyle\frac{2020}{2021}+\left(\displaystyle\frac{2}{5}-\displaystyle\frac{2}{7}\right):\displaystyle\frac{2020}{2021}\)

\(=\left[\left(-\displaystyle\frac{5}{7}\right)+\displaystyle\frac{3}{5}\right].\displaystyle\frac{2021}{2020}+\left(\displaystyle\frac{2}{5}-\displaystyle\frac{2}{7}\right).\displaystyle\frac{2021}{2020}\)

\(=\left[\left(-\displaystyle\frac{5}{7}\right)+\displaystyle\frac{3}{5}+\displaystyle\frac{2}{5}-\displaystyle\frac{2}{7}\right]+.\displaystyle\frac{2021}{2020}\)

\(=\left[\left(-\displaystyle\frac{5}{7}-\displaystyle\frac{2}{7}\right)+\left(\displaystyle\frac{3}{5}+\displaystyle\frac{2}{5}\right)\right]+.\displaystyle\frac{2021}{2020}\)

\(=(1-1).\displaystyle\frac{2021}{2020}=0.\)

d) \(\displaystyle\frac{3}{8}:\left(\displaystyle\frac{7}{22}-\displaystyle\frac{2}{11}\right)+\displaystyle\frac{3}{8}:\left(\displaystyle\frac{2}{5}-\displaystyle\frac{1}{10}\right)\)

\(=\displaystyle\frac{3}{8}:\left(\displaystyle\frac{7}{22}-\displaystyle\frac{4}{22}\right)+\displaystyle\frac{3}{8}:\left(\displaystyle\frac{4}{10}-\displaystyle\frac{1}{10}\right)\)

\(=\displaystyle\frac{3}{8}:\displaystyle\frac{3}{22}+\displaystyle\frac{3}{8}:\displaystyle\frac{3}{10}=\displaystyle\frac{3}{8}.\displaystyle\frac{22}{3}+\displaystyle\frac{3}{8}.\displaystyle\frac{10}{3}\)

\(=\displaystyle\frac{3}{8}.\left(\displaystyle\frac{22}{3}+\displaystyle\frac{10}{3}\right)=\displaystyle\frac{3}{8}.\displaystyle\frac{32}{3}=4.\)

\(\)

Xem thêm các bài giải khác tại: Giải Bài tập Toán Lớp 7 – NXB Chân Trời Sáng Tạo

Thông tin liên hệ & mạng xã hội:

Website: https://bumbii.com/

Diễn đàn hỏi đáp: https://hoidap.bumbii.com

Facebook: https://www.facebook.com/bumbiihome

Pinterest: https://www.pinterest.com/bumbiihome/

0 0 đánh giá
Article Rating
guest

0 Bình luận
Phản hồi nội tuyến
Xem tất cả bình luận
0
Rất thích suy nghĩ của bạn, hãy bình luận.x