Giải Toán 7 trang 15 Tập 1 Chân trời sáng tạo

Giải Toán 7 trang 15 Tập 1

Bài 1 trang 15 Toán lớp 7 Tập 1: Tính:

a) $\frac{2}{15}+\left(\frac{-5}{24}\right);$       

b) $\left(\frac{-5}{9}\right)-\left(-\frac{7}{27}\right);$   

c) $\left(\frac{-7}{12}\right)+0,75;$

d) $\left(\frac{-5}{9}\right)-1,25;$     

e) $0,34\cdot \left(\frac{-5}{17}\right);$     

g) $\frac{4}{9}:\left(\frac{-8}{15}\right);$

h) $\left(1\frac{2}{3}\right):\left(2\frac{1}{2}\right);$     

i) $\frac{2}{5}\cdot \left(-1,25\right);$   

k) $\left(\frac{-3}{5}\right)\cdot \left(\frac{15}{-7}\right)\cdot 3\frac{1}{9};$

Lời giải:

a)$\frac{2}{15}+\left(\frac{-5}{24}\right)=\frac{16}{120}+\left(\frac{-25}{120}\right)$$=\frac{-9}{120}=\frac{-3}{40}$

b) $\left(\frac{-5}{9}\right)-\left(-\frac{7}{27}\right)=\left(\frac{-15}{27}\right)+\frac{7}{27}=\frac{(-15)+7}{27}=\frac{-8}{27}.$

c) $\left(\frac{-7}{12}\right)+0,75=\left(\frac{-7}{12}\right)+\frac{75}{100}$

$=\left(\frac{-7}{12}\right)+\frac{3}{4}=\left(\frac{-7}{12}\right)+\frac{9}{12}=\frac{2}{12}=\frac{1}{6}.$

d) $\left(\frac{-5}{9}\right)-1,25=\left(\frac{-5}{9}\right)-\frac{125}{100}$

$=\left(\frac{-5}{9}\right)-\frac{5}{4}=\left(\frac{-20}{36}\right)-\frac{45}{36}=\frac{-65}{36}.$

e) $0,34\cdot \left(\frac{-5}{17}\right)=\frac{34}{100}\cdot \left(\frac{-5}{17}\right)$

$=\frac{17}{50}\cdot \left(\frac{-5}{17}\right)=\frac{-5}{50}=\frac{-1}{10}.$

g) $\frac{4}{9}:\left(\frac{-8}{15}\right)=\frac{4}{9}\cdot \left(\frac{-15}{8}\right)$

$=\frac{4.\left(-15\right)}{9.8}=\frac{-60}{72}=\frac{-5}{6}.$

h) $\left(1\frac{2}{3}\right):\left(2\frac{1}{2}\right)=\frac{5}{3}:\frac{5}{2}=\frac{5}{3}\cdot \frac{2}{5}=\frac{2}{3}.$

i) $\frac{2}{5}\cdot \left(-1,25\right)=\frac{2}{5}\cdot \frac{-125}{100}=\frac{2}{5}\cdot \frac{-5}{4}=\frac{-1}{2}.$

k) $\left(\frac{-3}{5}\right)\cdot \left(\frac{15}{-7}\right)\cdot 3\frac{1}{9}=\left(\frac{-3}{5}\right)\cdot \left(\frac{15}{-7}\right)\cdot \frac{28}{9}$

$=\frac{(-3)\mathrm{.15.28}}{5.(-7).9}=\frac{(-3)\mathrm{.3.5.7.4}}{5.(-7).9}=4.$

Bài 2 trang 15 Toán lớp 7 Tập 1: Tính:

a) $0,75-\frac{5}{6}+1\frac{1}{2};$

b)$\frac{3}{7}+\frac{4}{15}+\left(\frac{-8}{21}\right)+(-0,4);$

c)$0,625+\left(\frac{-2}{7}\right)+\frac{3}{8}+\left(\frac{-5}{7}\right)+1\frac{2}{3};$

d)$(-3)\cdot \left(\frac{-38}{21}\right)\cdot \left(\frac{-7}{6}\right)\cdot \left(-\frac{3}{19}\right);$

e)$\left(\frac{11}{18}:\frac{22}{9}\right)\cdot \frac{8}{5};$ 

g)$\left[\left(\frac{-4}{5}\right)\cdot \frac{5}{8}\right]:\left(\frac{-25}{12}\right).$

Lời giải:

a) $0,75-\frac{5}{6}+1\frac{1}{2}=\frac{75}{100}-\frac{5}{6}+\frac{3}{2}$

$=\frac{3}{4}-\frac{5}{6}+\frac{3}{2}$

$=\frac{9}{12}-\frac{10}{12}+\frac{18}{12}=\frac{17}{12}.$

b) $\frac{3}{7}+\frac{4}{15}+\left(\frac{-8}{21}\right)+(-0,4)$

$=\frac{3}{7}+\left(\frac{-8}{21}\right)+\frac{4}{15}+\left(\frac{-4}{10}\right)$

$=\left(\frac{9}{21}+\frac{-8}{21}\right)+\frac{4}{15}+\frac{-2}{5}$

$=\frac{1}{21}+\left(\frac{4}{15}+\frac{-6}{15}\right)$

$=\frac{1}{21}+\frac{-2}{15}=\frac{5}{105}+\frac{-14}{105}=\frac{-9}{105}=\frac{-3}{35}.$

c) $0,625+\left(\frac{-2}{7}\right)+\frac{3}{8}+\left(\frac{-5}{7}\right)+1\frac{2}{3}$

$=\frac{625}{1000}+\left(\frac{-2}{7}\right)+\frac{3}{8}+\left(\frac{-5}{7}\right)+\frac{5}{3}$

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

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

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

d) $(-3)\cdot \left(\frac{-38}{21}\right)\cdot \left(\frac{-7}{6}\right)\cdot \left(-\frac{3}{19}\right)$

$=\frac{(-3).(-38).(-7).(-3)}{\mathrm{21.6.19}}$

$=\frac{(-3).(-7).(-38).(-3)}{\mathrm{21.6.19}}$

$=\frac{\mathrm{3.7.2.19.3}}{\mathrm{3.7.2.3.19}}=1.$

e) $\left(\frac{11}{18}:\frac{22}{9}\right)\cdot \frac{8}{5}=\frac{11}{18}\cdot \frac{9}{22}\cdot \frac{8}{5}$

$=\frac{\mathrm{11.9.4.2}}{\mathrm{9.2.11.2.5}}=\frac{2}{5}.$

g) $\left[\left(\frac{-4}{5}\right)\cdot \frac{5}{8}\right]:\left(\frac{-25}{12}\right)$

$=\frac{\left(-4\right).5}{5.8}.\left(\frac{-12}{25}\right)$

$=\frac{-1}{2}.\frac{-12}{25}=\frac{12}{50}=\frac{6}{25}.$

Bài 3 trang 15 Toán lớp 7 Tập 1: Thay $\boxed{?}$ bằng dấu (>,<,=) thích hợp.

a) $\left(\frac{-5}{8}\right)+\left(\frac{3}{-8}\right)\boxed{?}$ -1 ;

b)$\frac{-8}{11}\boxed{?}\left(\frac{-13}{22}\right)+\left(\frac{-5}{22}\right);$

c)$\frac{1}{6}+\left(\frac{-3}{4}\right)\boxed{?}$ $\frac{1}{14}+\left(\frac{-4}{7}\right).$

Lời giải:

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

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

b) $\left(\frac{-13}{22}\right)+\left(\frac{-5}{22}\right)=\frac{-18}{22}=\frac{-9}{11}$

Do –8 > – 9 nên $\frac{-8}{11}>\frac{-9}{11}$

Vậy $\frac{-8}{11}\boxed{>}\left(\frac{-13}{22}\right)+\left(\frac{-5}{22}\right).$

c) $\frac{1}{6}+\left(\frac{-3}{4}\right)=\frac{2}{12}+\left(\frac{-9}{12}\right)=\frac{-7}{12}=\frac{-49}{84}.$

$\frac{1}{14}+\left(\frac{-4}{7}\right)=\frac{1}{14}+\left(\frac{-8}{14}\right)=\frac{-7}{14}=\frac{-42}{84}.$

Ta có –49 < –42 nên $\frac{-49}{84}<\frac{-42}{84}$ hay $\frac{-7}{12}<\frac{-7}{14}.$

Vậy $\frac{1}{6}+\left(\frac{-3}{4}\right)\boxed{<}\frac{1}{14}+\left(\frac{-4}{7}\right).$

Bài 4 trang 15 Toán lớp 7 Tập 1: Tính:

a)$\frac{3}{7}\cdot \left(\frac{-1}{9}\right)+\frac{3}{7}\cdot \left(-\frac{2}{3}\right);$

b)$\left(\frac{-7}{13}\right)\cdot \frac{5}{12}+\left(\frac{-7}{13}\right)\cdot \frac{7}{12}+\left(\frac{-6}{13}\right);$

c)$\left[\left(\frac{-2}{3}\right)+\frac{3}{7}\right]:\frac{5}{9}+\left(\frac{4}{7}-\frac{1}{3}\right):\frac{5}{9};$

d)$\frac{5}{9}:\left(\frac{1}{11}-\frac{5}{22}\right)+\frac{5}{9}:\left(\frac{1}{15}-\frac{2}{3}\right);$

e)

Lời giải:

a)$\frac{3}{7}\cdot \left(\frac{-1}{9}\right)+\frac{3}{7}\cdot \left(-\frac{2}{3}\right)$$=\frac{3}{7}\cdot \left[\left(\frac{-1}{9}\right)+\left(-\frac{2}{3}\right)\right]$

$=\frac{3}{7}\cdot \left[\left(\frac{-1}{9}\right)+\left(\frac{-6}{9}\right)\right]$

$=\frac{3}{7}\cdot \left(\frac{-7}{9}\right)=\frac{-1}{3}$

b)$\left(\frac{-7}{13}\right)\cdot \frac{5}{12}+\left(\frac{-7}{13}\right)\cdot \frac{7}{12}+\left(\frac{-6}{13}\right)$

$=\left(\frac{-7}{13}\right)\cdot \left(\frac{5}{12}+\frac{7}{12}\right)+\left(\frac{-6}{13}\right)$

$=\left(\frac{-7}{13}\right)\cdot \frac{12}{12}+\left(\frac{-6}{13}\right)$

$=\left(\frac{-7}{13}\right)+\left(\frac{-6}{13}\right)$$=\frac{-13}{13}=-1$

c)$\left[\left(\frac{-2}{3}\right)+\frac{3}{7}\right]:\frac{5}{9}+\left(\frac{4}{7}-\frac{1}{3}\right):\frac{5}{9}$

$=\left[\left(\frac{-2}{3}\right)+\frac{3}{7}\right]\cdot \frac{9}{5}+\left(\frac{4}{7}-\frac{1}{3}\right)\cdot \frac{9}{5}$

$=\frac{9}{5}\cdot \left[\left(\frac{-2}{3}\right)+\frac{3}{7}+\frac{4}{7}-\frac{1}{3}\right]$

$=\frac{9}{5}\cdot \left[\left(\frac{-2}{3}\right)-\frac{1}{3}+\frac{3}{7}+\frac{4}{7}\right]$

$=\frac{9}{5}\cdot (-1+1)=\frac{9}{5}.0=0.$

d) $\frac{5}{9}:\left(\frac{1}{11}-\frac{5}{22}\right)+\frac{5}{9}:\left(\frac{1}{15}-\frac{2}{3}\right)$

$=\frac{5}{9}:\left(\frac{2}{22}-\frac{5}{22}\right)+\frac{5}{9}:\left(\frac{1}{15}-\frac{10}{15}\right)$

$=\frac{5}{9}:\left(\frac{-3}{22}\right)+\frac{5}{9}:\left(\frac{-9}{15}\right)=\frac{5}{9}\cdot \frac{-22}{3}+\frac{5}{9}\cdot \frac{-5}{3}$

$=\frac{5}{9}\cdot \left(\frac{-22}{3}+\frac{-5}{3}\right)=\frac{5}{9}.\frac{-27}{3}=\frac{5}{9}\cdot (-9)=-5.$

e) $\frac{3}{5}+\frac{3}{11}-\left(\frac{-3}{7}\right)+\left(\frac{-2}{97}\right)-\frac{1}{35}-\frac{3}{4}+\left(\frac{-23}{44}\right)$

$=\left[\frac{3}{5}-\left(\frac{-3}{7}\right)-\frac{1}{35}\right]+\left[\frac{3}{11}-\frac{3}{4}+\left(\frac{-23}{44}\right)\right]+\left(\frac{-2}{97}\right)$

$=\left(\frac{21}{35}+\frac{15}{35}-\frac{1}{35}\right)+\left(\frac{12}{44}-\frac{33}{44}-\frac{23}{44}\right)+\left(\frac{-2}{97}\right)$

$=\frac{35}{35}+\left(\frac{-44}{44}\right)+\left(\frac{-2}{97}\right)=1+\left(-1\right)+\left(\frac{-2}{97}\right)=\frac{-2}{97}.$

Bài 5 trang 15 Toán lớp 7 Tập 1: Tìm x, biết:

a)$x\cdot \frac{14}{27}=\frac{-7}{9};$

b)$\left(\frac{-5}{9}\right):x=\frac{2}{3};$

c)$\frac{2}{5}:x=\frac{1}{16}:0,125;$

d)$-\frac{5}{12}x=\frac{2}{3}-\frac{1}{2}.$

Lời giải:

a)$x\cdot \frac{14}{27}=\frac{-7}{9};$                

$x=\frac{-7}{9}:\frac{14}{27}$

$x=\frac{-7}{9}\cdot \frac{27}{14}$

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

Vậy $x=\frac{-3}{2}$

b)$\left(\frac{-5}{9}\right):x=\frac{2}{3}$      

$x=\left(-\frac{5}{9}\right):\frac{2}{3}$        

$x=\left(-\frac{5}{9}\right)\cdot \frac{3}{2}$

$x=\frac{-5}{6}$        

Vậy $x=\frac{-5}{6}$

c)$\frac{2}{5}:x=\frac{1}{16}:0,125$

$\frac{2}{5}:x=\frac{1}{16}:\frac{125}{1000}$

$\frac{2}{5}:x=\frac{1}{16}:\frac{1}{8}$

$\frac{2}{5}:x=\frac{1}{16}\cdot 8$

$x=\frac{2}{5}:\frac{1}{2}$

x = $\frac{2}{5}\cdot 2$

x = $\frac{4}{5}$

Vậy x = $\frac{4}{5}$

d)$-\frac{5}{12}x=\frac{2}{3}-\frac{1}{2}.$

$\frac{-5}{12}x=\frac{4}{6}-\frac{3}{6}$

$\frac{-5}{12}x=\frac{1}{6}$

$x=\frac{1}{6}:\frac{-5}{12}$

$x=\frac{1}{6}\cdot \frac{12}{-5}$

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

Vậy $x=\frac{-2}{5}$

Xem thêm lời giải bài tập Toán lớp 7 Chân trời sáng tạo hay, chi tiết khác:

Giải Toán 7 trang 11 Tập 1

Giải Toán 7 trang 12 Tập 1

Giải Toán 7 trang 13 Tập 1

Giải Toán 7 trang 14 Tập 1

Giải Toán 7 trang 15 Tập 1

Giải Toán 7 trang 16 Tập 1

Xem thêm lời giải bài tập Toán lớp 7 Chân trời sáng tạo hay, chi tiết khác:

Bài 1: Tập hợp các số hữu tỉ

Bài 3: Lũy thừa của một số hữu tỉ

Bài 4: Quy tắc dấu ngoặc và quy tắc chuyển vế

Bài 5: Hoạt động thực hành và trải nghiệm: Thực hành tính tiền điện

Bài tập cuối chương 1