Giải SBT Toán 7 trang 10 Tập 1 Chân trời sáng tạo

Giải SBT Toán 7 trang 10 Tập 1 Chân trời sáng tạo

Bài 1 trang 10 Sách bài tập Toán 7 Tập 1: Tính:

a) $\frac{3}{10}+\left(\frac{-5}{12}\right)$;

b) $\frac{-3}{8}-\left(-\frac{7}{24}\right)$;

c) $\left(\frac{-5}{14}\right)+0,25$;

d) $\left(-\frac{7}{8}\right)-1,25$;

e) $\left(\frac{-5}{14}\right).\frac{21}{25}$;

f) $\frac{8}{27}:\left(-\frac{16}{45}\right)$;

g) $\left(1\frac{5}{6}\right):\left(4\frac{1}{8}\right)$;

h) $0,38.\left(\frac{-7}{19}\right)$;

i) $\left(\frac{-4}{5}\right).\left(\frac{15}{-8}\right).1\frac{1}{9}$.

Lời giải:

a) $\frac{3}{10}+\left(\frac{-5}{12}\right)=\frac{18}{60}+\left(\frac{-25}{60}\right)=\frac{-7}{60}$;

b) $\frac{-3}{8}-\left(-\frac{7}{24}\right)=\frac{-3}{8}+\frac{7}{24}$

$=\frac{-9}{24}+\frac{7}{24}=\frac{-2}{24}=\frac{-1}{12}$;

c) $\left(\frac{-5}{14}\right)+0,25=\left(\frac{-5}{14}\right)+\frac{1}{4}$

$=\left(\frac{-10}{28}\right)+\frac{7}{28}=\frac{-3}{28}$;

d) $\left(-\frac{7}{8}\right)-1,25=-\left(\frac{7}{8}+\frac{5}{4}\right)$

$=-\left(\frac{7}{8}+\frac{10}{8}\right)=\frac{-17}{8}$;

e) $\left(\frac{-5}{14}\right).\frac{21}{25}=-\frac{5.21}{14.25}$

$=-\frac{5.7.3}{7.2.5.5}=\frac{-3}{10}$;

f) $\frac{8}{27}:\left(-\frac{16}{45}\right)=\frac{8}{27}.\left(\frac{-45}{16}\right)$

$=-\frac{8.45}{27.16}=-\frac{8.9.5}{9.3.8.2}=\frac{-5}{6}$;

g) $\left(1\frac{5}{6}\right):\left(4\frac{1}{8}\right)=\frac{11}{6}:\frac{33}{8}$

$=\frac{11}{6}.\frac{8}{33}=\frac{11.2.4}{2.3.11.3}=\frac{4}{9}$;

h) $0,38.\left(\frac{-7}{19}\right)=\frac{19}{50}.\left(\frac{-7}{19}\right)$

$=\frac{19.(-7)}{50.19}=\frac{-7}{50}$;

i) $\left(\frac{-4}{5}\right).\left(\frac{15}{-8}\right).1\frac{1}{9}=\frac{4}{5}.\frac{15}{8}.\frac{10}{9}$

$=\frac{4.15.10}{5.8.9}=\frac{4.5.3.5.2}{5.4.2.3.3}=\frac{5}{3}$.

Bài 2 trang 10 Sách bài tập Toán 7 Tập 1: Tính:

a) $-0,375+\frac{1}{4}-\left(-\frac{1}{5}\right)$;

b) $\frac{2}{3}+\left(\frac{-5}{12}\right)+\left(\frac{-7}{15}\right)+(-0,2)$;

c) $0,275+\left(\frac{-8}{17}\right)+\frac{29}{40}+\left(\frac{-9}{17}\right)-1\frac{1}{3}$;

d) $(-5).\left(\frac{-34}{21}\right).\left(\frac{-7}{10}\right).\left(-\frac{3}{17}\right)$;

e) $\left(\frac{13}{18}:\frac{26}{9}\right).\frac{4}{7}$;

f) $\left[\left(-\frac{5}{33}\right):\frac{2}{11}\right].0,15$.

Lời giải:

a) $-0,375+\frac{1}{4}-\left(-\frac{1}{5}\right)=\frac{-3}{8}+\frac{1}{4}+\frac{1}{5}$

$=\frac{-15}{40}+\frac{10}{40}+\frac{8}{40}=\frac{3}{40}$;

b) $\frac{2}{3}+\left(\frac{-5}{12}\right)+\left(\frac{-7}{15}\right)+(-0,2)$

$=\frac{2}{3}-\frac{5}{12}-\frac{7}{15}-\frac{1}{5}$

$=\left(\frac{2}{3}-\frac{5}{12}\right)-\left(\frac{7}{15}+\frac{1}{5}\right)$

$=\left(\frac{8}{12}-\frac{5}{12}\right)-\left(\frac{7}{15}+\frac{3}{15}\right)$

$=\frac{3}{12}-\frac{10}{15}=\frac{3}{12}-\frac{8}{12}=\frac{-5}{12}$;

c) $0,275+\left(\frac{-8}{17}\right)+\frac{29}{40}+\left(\frac{-9}{17}\right)-1\frac{1}{3}$

$=\frac{11}{40}+\left(\frac{-8}{17}\right)+\frac{29}{40}+\left(\frac{-9}{17}\right)-\frac{4}{3}$

$=\left(\frac{11}{40}+\frac{29}{40}\right)+\left[\left(\frac{-8}{17}\right)+\left(\frac{-9}{17}\right)\right]-\frac{4}{3}$

$=1+(-1)-\frac{4}{3}=\frac{-4}{3}$;

d) $(-5).\left(\frac{-34}{21}\right).\left(\frac{-7}{10}\right).\left(-\frac{3}{17}\right)$

$=5.\frac{34}{21}.\frac{7}{10}.\frac{3}{17}=\frac{5.34.7.3}{21.10.17}$

$=\frac{5.17.2.7.3}{7.3.5.2.17}=1$;

e) $\left(\frac{13}{18}:\frac{26}{9}\right).\frac{4}{7}=\frac{13}{18}.\frac{9}{26}.\frac{4}{7}$

$=\frac{13}{18}.\frac{9}{26}.\frac{4}{7}=\frac{13.9.4}{18.26.7}$

$=\frac{13.9.2.2}{9.2.13.2.7}=\frac{1}{7}$;

f) $\left[\left(-\frac{5}{33}\right):\frac{2}{11}\right].0,15=\frac{-5}{33}.\frac{11}{2}.\frac{3}{20}$

$=-\frac{5.11.3}{33.2.20}=-\frac{5.11.3}{11.3.2.5.4}=\frac{-1}{8}$.

Bài 3 trang 10 Sách bài tập Toán 7 Tập 1: Thay dấu $\boxed{?}$ bằng dấu >, <, =  thích hợp.

a) $\left(\frac{-4}{9}\right)+\left(\frac{5}{-9}\right)\boxed{?}-1$;

b) $\left(\frac{-13}{33}\right)+\left(\frac{-14}{33}\right)\boxed{?}\left(\frac{-10}{11}\right)$;

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

Lời giải:

a) Ta có: $\left(\frac{-4}{9}\right)+\left(\frac{5}{-9}\right)=\left(\frac{-4}{9}\right)+\left(\frac{-5}{9}\right)=\frac{-9}{9}=-1$

Vậy $\left(\frac{-4}{9}\right)+\left(\frac{5}{-9}\right)\boxed{=}-1$;

b) Ta có: $\left(\frac{-13}{33}\right)+\left(\frac{-14}{33}\right)=\frac{-27}{33}=\frac{-9}{11}$.

Vì −9 > −10 nên $\frac{-9}{11}>\frac{-10}{11}$.

Vậy $\left(\frac{-13}{33}\right)+\left(\frac{-14}{33}\right)\boxed{>}\left(\frac{-10}{11}\right)$;

c) Ta có: $\frac{1}{5}+\left(\frac{-2}{3}\right)=\frac{3}{15}+\left(\frac{-10}{15}\right)=\frac{-7}{15}$;

$\frac{2}{5}+\left(\frac{-4}{5}\right)=\frac{-2}{5}=\frac{-6}{15}$.

Vì −7 < −6 nên $\frac{-7}{15}<\frac{-6}{15}$.

Vậy $\frac{1}{5}+\left(\frac{-2}{3}\right)\boxed{<}\frac{2}{5}+\left(\frac{-4}{5}\right)$.

Bài 4 trang 10 Sách bài tập Toán 7 Tập 1:

a) $\frac{6}{7}.\left(\frac{-1}{8}\right)+\frac{6}{7}.\left(-\frac{3}{4}\right)$;

b) $\left(\frac{-7}{17}\right).\frac{5}{12}+\left(\frac{-7}{17}\right).\frac{7}{12}+\left(\frac{-10}{17}\right)$;

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

d) $\frac{7}{8}:\left(\frac{2}{9}-\frac{1}{18}\right)+\frac{7}{8}:\left(\frac{1}{36}-\frac{5}{12}\right)$.

Lời giải:

a) $\frac{6}{7}.\left(\frac{-1}{8}\right)+\frac{6}{7}.\left(-\frac{3}{4}\right)=\frac{6}{7}.\left[\left(\frac{-1}{8}\right)+\left(\frac{-3}{4}\right)\right]$

$=\frac{6}{7}.\left[\left(\frac{-1}{8}\right)+\left(\frac{-6}{8}\right)\right]=\frac{6}{7}.\frac{-7}{8}=\frac{-3}{4}$;

b) $\left(\frac{-7}{17}\right).\frac{5}{12}+\left(\frac{-7}{17}\right).\frac{7}{12}+\left(\frac{-10}{17}\right)$

$=\left(\frac{-7}{17}\right).\left(\frac{5}{12}+\frac{7}{12}\right)+\left(\frac{-10}{17}\right)$

$=\left(\frac{-7}{17}\right).1+\left(\frac{-10}{17}\right)=\left(\frac{-7}{17}\right)+\left(\frac{-10}{17}\right)$

$=\frac{-17}{17}=-1$;

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

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

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

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

$=\left[1+\left(-1\right)\right].\frac{7}{3}=0.\frac{7}{3}=0$;

d) $\frac{7}{8}:\left(\frac{2}{9}-\frac{1}{18}\right)+\frac{7}{8}:\left(\frac{1}{36}-\frac{5}{12}\right)$

$=\frac{7}{8}:\left(\frac{4}{18}-\frac{1}{18}\right)+\frac{7}{8}:\left(\frac{1}{36}-\frac{15}{36}\right)$

$=\frac{7}{8}:\frac{1}{6}+\frac{7}{8}:\left(\frac{-7}{18}\right)$

$=\frac{7}{8}.6+\frac{7}{8}.\left(\frac{-18}{7}\right)$

$=\frac{7}{8}.\left[6+\left(\frac{-18}{7}\right)\right]$

$=\frac{7}{8}.\left[\frac{42}{7}+\left(\frac{-18}{7}\right)\right]$

$=\frac{7}{8}.\frac{24}{7}=3$.

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