Tính Bài 2 trang 10 sách bài tập Toán lớp 7 Tập 1

Tính:

Giải sách bài tập Toán lớp 7 Bài 2: Các phép tính với số hữu tỉ

Bài 2 trang 10 sách bài tập Toán lớp 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) $\left[\left(-\frac{5}{33}\right):\frac{2}{11}\right].0,15=\frac{-5}{33}.\frac{11}{2}.\frac{3}{20}$

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