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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