Sách bài tập Toán 7 Bài 4 (Chân trời sáng tạo): Quy tắc dấu ngoặc và quy tắc chuyển vế

Giải sách bài tập Toán lớp 7 Bài 4: Quy tắc dấu ngoặc và quy tắc chuyển vế - Chân trời sáng tạo

Giải SBT Toán 7 trang 17 Tập 1

Bài 1 trang 17Sách bài tập Toán 7 Tập 1: Bỏ ngoặc rồi tính.

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

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

c) $\left.\left(\frac{-2}{5}\right)+\frac{1}{3}\right]-\left(\frac{3}{5}-\frac{1}{4}\right)$;

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

Lời giải:

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

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

$=\frac{-9}{9}+\frac{7}{9}=\frac{-2}{9}$;

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

$=\frac{4}{9}-\frac{2}{9}-\frac{3}{7}=\frac{2}{9}-\frac{3}{7}$

$=\frac{14}{63}-\frac{27}{63}=\frac{-13}{63}$;

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

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

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

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

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

= 1 – 1 = 0.

Giải SBT Toán 7 trang 18 Tập 1

Bài 2 trang 18Sách bài tập Toán 7 Tập 1: Tính:

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

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

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

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

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

Lời giải:

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

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

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

$=\left(\frac{-1}{2}\right)+\frac{2}{9}+\left(\frac{-1}{4}\right)$

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

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

$=\left.\left(\frac{-7}{8}\right).\frac{16}{21}\right]-\frac{5}{3}.\left(\frac{10}{30}-\frac{21}{30}\right)$

$=\frac{(-7).16}{8.21}-\frac{5}{3}.\left(\frac{-11}{30}\right)$

$=-\frac{7.8.2}{8.7.3}+\frac{5.11}{3.5.6}$

$=\frac{-2}{3}+\frac{11}{18}=\frac{-1}{18}$;

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

$={\left.\left(\frac{-8}{12}\right)+\frac{9}{12}\right]}^2.\frac{12}{5}-\frac{1}{5}$

$={\left(\frac{1}{12}\right)}^2.\frac{12}{5}-\frac{1}{5}$

$=\frac{1}{{12}^2}.\frac{12}{5}-\frac{1}{5}$

$=\frac{1}{60}-\frac{1}{5}=\frac{-11}{60}$;

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

$={\left(\frac{1}{25}-\frac{2}{5}\right)}^2.\frac{125}{9}-\left.\left(\frac{4}{3}-\frac{2}{5}\right).\frac{3}{7}\right]$

$={\left(\frac{-9}{25}\right)}^2.\frac{125}{9}-\left.\frac{14}{15}.\frac{3}{7}\right]$

$=\frac{9^2}{{25}^2}.\frac{125}{9}-\frac{2}{5}=\frac{{\left(3^2\right)}^2}{{\left(5^2\right)}^2}.\frac{5^3}{3^2}-\frac{2}{5}$

$=\frac{3^4}{5^4}.\frac{5^3}{3^2}-\frac{2}{5}=\frac{3^2}{5}-\frac{2}{5}$

$=\frac{9}{5}-\frac{2}{5}=\frac{7}{5}$;

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

$=\left.\frac{71}{18}.\left.\frac{5}{2}-\frac{5}{9}\right]\right\}:{\left.\left(\frac{-1}{2}\right)+\frac{1}{4}\right]}^2$

$=\left.\frac{71}{18}.\frac{35}{18}\right\}:{\left.\frac{-1}{4}\right]}^2=\frac{2485}{324}:\frac{1}{16}$

$=\frac{2485}{324}.16=\frac{9940}{81}$.

Bài 3 trang 18 Sách bài tập Toán 7 Tập 1: Cho biểu thức:

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

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.

Lời giải:

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

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

$=\frac{47}{6}-\frac{7}{6}-\frac{49}{6}=\frac{-9}{6}=\frac{-3}{2}$.

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

$=8-\frac{2}{3}+\frac{1}{2}-5+\frac{7}{3}+\frac{3}{2}-\frac{5}{3}-\frac{5}{2}-4$

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

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

Bài 4 trang 18 Sách bài tập Toán 7 Tập 1: Tìm x, biết:

a) $x+\frac{3}{7}=\frac{2}{5}$;

b) $\frac{3}{2}-x=\frac{4}{5}$;

c) $\frac{5}{9}-\frac{1}{3}x=\frac{2}{3}$;

d) $\frac{3}{5}x-1\frac{1}{5}=\frac{-3}{14}:\frac{5}{7}$.

Lời giải:

a) $x+\frac{3}{7}=\frac{2}{5}$

$x=\frac{2}{5}-\frac{3}{7}$

$x=\frac{14}{35}-\frac{15}{35}$

$x=\frac{-1}{35}$.

Vậy $x=\frac{-1}{35}$.

b) $\frac{3}{2}-x=\frac{4}{5}$

$x=\frac{3}{2}-\frac{4}{5}$

$x=\frac{15}{10}-\frac{8}{10}$

$x=\frac{7}{10}$.

Vậy $x=\frac{7}{10}$.

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

$\frac{1}{3}x=\frac{5}{9}-\frac{2}{3}$

$\frac{1}{3}x=\frac{-1}{9}$

$x=\left(\frac{-1}{9}\right):\frac{1}{3}$

$x=\frac{-1}{3}$.

Vậy $x=\frac{-1}{3}$.

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

$\frac{3}{55}x-\frac{6}{5}=\frac{-3}{10}$

$\frac{3}{5}x=\frac{-3}{10}+\frac{6}{5}$

$\frac{3}{5}x=\frac{9}{10}$

$x=\frac{9}{10}:\frac{3}{5}$

$x=\frac{3}{2}$.

Vậy $x=\frac{3}{2}$.

Bài 5 trang 18 Sách bài tập Toán 7 Tập 1:Tìm x, biết:

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

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

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

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

Lời giải:

a) $\frac{3}{4}+\frac{1}{9}:x=0,5$

$\frac{1}{9}:x=\frac{1}{2}-\frac{3}{4}$

$\frac{1}{9}:x=\frac{-1}{4}$

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

$x=\frac{1}{9}.(-4)$

$x=\frac{-4}{9}$.

Vậy $x=\frac{-4}{9}$.

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

$x-\frac{2}{3}=\frac{3}{4}-\frac{4}{3}$

$x-\frac{2}{3}=\frac{-7}{12}$

$x=\frac{-7}{12}+\frac{2}{3}$

$x=\frac{1}{12}$.

Vậy $x=\frac{1}{12}$.

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

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

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

$\frac{5}{7}-x=\frac{-2}{3}$

$x=\frac{5}{7}-\left(\frac{-2}{3}\right)$

$x=\frac{5}{7}+\frac{2}{3}$

$x=\frac{29}{21}$.

Vậy $x=\frac{29}{21}$.

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

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

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

$\frac{5}{2}x-\frac{4}{7}=\frac{-4}{7}$

$\frac{5}{2}x=\left(\frac{-4}{7}\right)+\frac{4}{7}$

$\frac{5}{2}x=0$

x = 0.

Vậy x = 0.

Bài 6 trang 18, 19 Sách bài tập Toán 7 Tập 1: Tính nhanh:

a) $\frac{12}{23}.\frac{7}{13}+\frac{11}{23}.\frac{7}{13}$;

b) $\frac{4}{9}.\frac{23}{11}-\frac{1}{11}.\frac{4}{9}+\frac{4}{9}$;

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

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

Lời giải:

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

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

$=\frac{7}{13}.1=\frac{7}{13}$.

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

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

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

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

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

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

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

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

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

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

$=\frac{3}{8}:\frac{3}{22}+\frac{3}{8}:\frac{3}{10}$$=\frac{3}{8}.\frac{22}{3}+\frac{3}{8}.\frac{10}{3}$

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