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 \(24\) sách giáo khoa toán lớp \(7\) tập \(1\) NXB Chân Trời Sáng Tạo.

\(1.\) Bỏ dấu ngoặc rồi tính.

a) \(\left(\displaystyle\frac{-3}{7}\right)+\left(\displaystyle\frac{5}{6}-\displaystyle\frac{4}{7}\right); \hspace{3,5cm} b) \displaystyle\frac{3}{5}-\left(\displaystyle\frac{2}{3}+\displaystyle\frac{1}{5}\right);\)

c) \(\left[\left(\displaystyle\frac{-1}{3}\right)+1\right]-\left(\displaystyle\frac{2}{3}-\displaystyle\frac{1}{5}\right); \hspace{2cm} d) 1\displaystyle\frac{1}{3}+\left(\displaystyle\frac{2}{3}-\displaystyle\frac{3}{4}\right)-\left(0,8+1\displaystyle\frac{1}{5}\right).\)

Giải

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

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

\(=\displaystyle\frac{5}{6}+(-1)=\displaystyle\frac{5}{6}+\displaystyle\frac{-6}{6}=\displaystyle\frac{-1}{6};\)

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

\(=\left(\displaystyle\frac{3}{5}-\displaystyle\frac{1}{5}\right)-\displaystyle\frac{2}{3}=\displaystyle\frac{2}{5}-\displaystyle\frac{2}{3}=\displaystyle\frac{6}{15}-\displaystyle\frac{10}{15}=\displaystyle\frac{-4}{15};\)

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

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

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

\(=\displaystyle\frac{4}{3}+\displaystyle\frac{2}{3}-\displaystyle\frac{3}{4}-\displaystyle\frac{8}{10}-\displaystyle\frac{6}{5}=\displaystyle\frac{4}{3}+\displaystyle\frac{2}{3}-\displaystyle\frac{3}{4}-\displaystyle\frac{4}{5}-\displaystyle\frac{6}{5}\)

\(=\displaystyle\frac{6}{3}-\displaystyle\frac{3}{4}-\displaystyle\frac{10}{5}=2-\displaystyle\frac{3}{4}-2=-\displaystyle\frac{3}{4}.\)

\(\)

\(2.\) Tính:

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

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

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

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

Giải

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

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

b) \(\left[\left(\displaystyle\frac{-1}{5}\right):\displaystyle\frac{1}{10}\right]-\displaystyle\frac{5}{7}.\left(\displaystyle\frac{2}{3}-\displaystyle\frac{1}{5}\right)=\left[\left(\displaystyle\frac{-1}{5}\right).10\right]-\displaystyle\frac{5}{7}.\left(\displaystyle\frac{10}{15}-\displaystyle\frac{3}{15}\right)\)

\(=-2-\displaystyle\frac{5}{7}.\displaystyle\frac{7}{15}=-2-\displaystyle\frac{1}{3}=\displaystyle\frac{-6}{3}-\displaystyle\frac{1}{3}=\displaystyle\frac{-7}{3}.\)

c) \((-0,4)+2\displaystyle\frac{2}{5}.\left[\left(\displaystyle\frac{-2}{3}\right)+\displaystyle\frac{1}{2}\right]^2=\displaystyle\frac{-4}{10}+\displaystyle\frac{12}{5}.\left(\displaystyle\frac{-4}{6}+\displaystyle\frac{3}{6}\right)^2\)

\(=\displaystyle\frac{-2}{5}+\displaystyle\frac{12}{5}.\left(\displaystyle\frac{-1}{6}\right)^2=\displaystyle\frac{-2}{5}+\displaystyle\frac{12}{5}.\displaystyle\frac{1}{36}\)

\(=\displaystyle\frac{-2}{5}+\displaystyle\frac{1}{15}=\displaystyle\frac{-6}{15}+\displaystyle\frac{1}{15}=\displaystyle\frac{-5}{15}=\displaystyle\frac{-1}{3}.\)

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

\(=\left\{\left[\left(\displaystyle\frac{1}{25}-\displaystyle\frac{3}{5}\right)^2:\displaystyle\frac{7^2}{5^3}\right].\displaystyle\frac{5}{6}\right\}-\left[\left(\displaystyle\frac{-2}{6}\right)+\displaystyle\frac{3}{6}\right]\)

\(=\left\{\left[\left(\displaystyle\frac{1}{25}-\displaystyle\frac{15}{25}\right)^2:\displaystyle\frac{7^2}{5^3}\right].\displaystyle\frac{5}{6}\right\}-\displaystyle\frac{1}{6}\)

\(=\left[\left(\displaystyle\frac{-14}{25}\right)^2.\displaystyle\frac{5^3}{7^2}\right].\displaystyle\frac{5}{6}-\displaystyle\frac{1}{6}\)

\(=\displaystyle\frac{14^2.5^3.5}{(5^2)^2.7^2.6}-\displaystyle\frac{1}{6}=\displaystyle\frac{(2.7)^2.5^4}{5^4.7^2.6}-\displaystyle\frac{1}{6}\)

\(=\displaystyle\frac{2^2.7^2.5^4}{5^4.7^2.6}-\displaystyle\frac{1}{6}=\displaystyle\frac{4}{6}-\displaystyle\frac{1}{6}=\displaystyle\frac{1}{2}.\)

\(\)

\(3.\) Cho biểu thức:

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

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.

Giải

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

\(A=\left(\displaystyle\frac{30}{15}+\displaystyle\frac{5}{15}-\displaystyle\frac{6}{15}\right)-\left(\displaystyle\frac{105}{15}-\displaystyle\frac{9}{15}-\displaystyle\frac{20}{15}\right)-\left(\displaystyle\frac{3}{15}+\displaystyle\frac{25}{15}-\displaystyle\frac{60}{15}\right)\)

\(A=\displaystyle\frac{29}{15}-\displaystyle\frac{76}{15}-\left(\displaystyle\frac{-32}{15}\right)=\displaystyle\frac{-15}{15}=-1.\)

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

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

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

\(A=-1+\displaystyle\frac{0}{3}-\displaystyle\frac{0}{5}=-1.\)

\(\)

\(4.\)Tìm x, biết:

\(a)\ x+\displaystyle\frac{3}{5}=\displaystyle\frac{2}{3}; \hspace{3,5cm} b)\ \displaystyle\frac{3}{7}-x=\displaystyle\frac{2}{5};\)

\(c)\ \displaystyle\frac{4}{9}-\displaystyle\frac{2}{3}x=\displaystyle\frac{1}{3}; \hspace{3cm} d)\ \displaystyle\frac{3}{10}x-1\displaystyle\frac{1}{2}=\left(\displaystyle\frac{-2}{7}\right):\displaystyle\frac{5}{14}.\)

Giải

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

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

\(x=\displaystyle\frac{10}{15}-\displaystyle\frac{9}{15}=\displaystyle\frac{1}{15}.\)

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

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

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

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

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

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

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

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

\(x=\displaystyle\frac{1}{9}.\displaystyle\frac{3}{2}=\displaystyle\frac{1}{6}\)

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

\(\displaystyle\frac{3}{10}x-\displaystyle\frac{3}{2}=\left(\displaystyle\frac{-2}{7}\right).\displaystyle\frac{14}{5}\)

\(\displaystyle\frac{3}{10}x-\displaystyle\frac{3}{2}=\displaystyle\frac{-4}{5}\)

\(\displaystyle\frac{3}{10}x=\displaystyle\frac{-4}{5}+\displaystyle\frac{3}{2}\)

\(\displaystyle\frac{3}{10}x=\displaystyle\frac{7}{10}\)

\(x=\displaystyle\frac{7}{10}:\displaystyle\frac{3}{10}\)

\(x=\displaystyle\frac{7}{10}.\displaystyle\frac{10}{3}=\displaystyle\frac{7}{3}.\)

\(\)

\(5.\) Tìm x, biết:

\(a)\ \displaystyle\frac{2}{9}:x+\displaystyle\frac{5}{6}=0,5; \hspace{4cm} b)\ \displaystyle\frac{3}{4}-\left(x-\displaystyle\frac{2}{3}\right)=1\displaystyle\frac{1}{3};\)

\(c)\ 1\displaystyle\frac{1}{4}:\left(x-\displaystyle\frac{2}{3}\right)=0,75; \hspace{3cm} d)\ \left(-\displaystyle\frac{5}{6}x+\displaystyle\frac{5}{4}\right):\displaystyle\frac{3}{2}=\displaystyle\frac{4}{3}.\)

Giải

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

\(\displaystyle\frac{2}{9}:x=\displaystyle\frac{1}{2}-\displaystyle\frac{5}{6}\)

\(\displaystyle\frac{2}{9}:x=\displaystyle\frac{3}{6}-\displaystyle\frac{5}{6}\)

\(\displaystyle\frac{2}{9}:x=\displaystyle\frac{-2}{6}\)

\(x=\displaystyle\frac{2}{9}:\left(\displaystyle\frac{-2}{6}\right)=\displaystyle\frac{-2}{3}.\)

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

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

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

\(-x=\displaystyle\frac{16}{12}-\displaystyle\frac{8}{12}-\displaystyle\frac{9}{12}\)

\(x=\displaystyle\frac{1}{12}.\)

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

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

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

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

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

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

\(x=\displaystyle\frac{7}{3}.\)

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

\(-\displaystyle\frac{5}{6}x+\displaystyle\frac{5}{4}=\displaystyle\frac{4}{3}.\displaystyle\frac{3}{2}\)

\(-\displaystyle\frac{5}{6}x+\displaystyle\frac{5}{4}=2\)

\(-\displaystyle\frac{5}{6}x=2-\displaystyle\frac{5}{4}\)

\(-\displaystyle\frac{5}{6}x=\displaystyle\frac{3}{4}\)

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

\(x=\displaystyle\frac{3}{4}.\displaystyle\frac{-6}{5}=\displaystyle\frac{-9}{10}.\)

\(\)

\(6.\) Tính nhanh.

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

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

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

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

Giải

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

\(=\displaystyle\frac{7}{11}.\left(\displaystyle\frac{13}{23}+\displaystyle\frac{10}{23}\right)\)

\(=\displaystyle\frac{7}{11}.1=\displaystyle\frac{7}{11}.\)

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

\(=\displaystyle\frac{5}{9}.\left(\displaystyle\frac{23}{11}-\displaystyle\frac{1}{11}+1\right)\)

\(=\displaystyle\frac{5}{9}.(2+1)=\displaystyle\frac{5}{9}.3=\displaystyle\frac{5}{3}.\)

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

\(=\left[\left(-\displaystyle\frac{4}{9}\right)+\displaystyle\frac{3}{5}\right].\displaystyle\frac{17}{13}+\left(\displaystyle\frac{2}{5}-\displaystyle\frac{5}{9}\right).\displaystyle\frac{17}{13}\)

\(=\left(-\displaystyle\frac{4}{9}+\displaystyle\frac{3}{5}+\displaystyle\frac{2}{5}-\displaystyle\frac{5}{9}\right).\displaystyle\frac{17}{13}\)

\(=\left[\left(-\displaystyle\frac{4}{9}-\displaystyle\frac{5}{9}\right)+\left(\displaystyle\frac{3}{5}+\displaystyle\frac{2}{5}\right)\right].\displaystyle\frac{17}{13}\)

\(=(-1+1).\displaystyle\frac{17}{13}=0.\)

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

\(=\displaystyle\frac{3}{16}:\left(\displaystyle\frac{3}{22}-\displaystyle\frac{6}{22}\right)+\displaystyle\frac{3}{16}:\left(\displaystyle\frac{1}{10}-\displaystyle\frac{4}{10}\right)\)

\(=\displaystyle\frac{3}{16}:\left(\displaystyle\frac{-3}{22}\right)+\displaystyle\frac{3}{16}:\left(\displaystyle\frac{-3}{10}\right)\)

\(=\displaystyle\frac{3}{16}.\left(\displaystyle\frac{-22}{3}\right)+\displaystyle\frac{3}{16}.\left(\displaystyle\frac{-10}{3}\right)\)

\(=\displaystyle\frac{3}{16}.\left(\displaystyle\frac{-22}{3}+\displaystyle\frac{-10}{3}\right)=\displaystyle\frac{3}{16}.\displaystyle\frac{-32}{3}=-2.\)

\(\)

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

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