Bài 4 trang 16 Toán 7 Tập 1 Chân trời sáng tạo

Giải Toán 7 Bài 2: Các phép tính với số hữu tỉ - Chân trời sáng tạo

Bài 4 trang 16 Toán 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) $\left[\left(\frac{-2}{3}\right)+\frac{3}{7}\right]:\frac{5}{9}+\left(\frac{4}{7}-\frac{1}{3}\right):\frac{5}{9}$

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

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

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

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

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

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

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

$=\frac{5}{9}\cdot \frac{-22}{3}+\frac{5}{9}\cdot \frac{-15}{9}$

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

$=\frac{5}{9}\cdot \frac{-27}{3}$ = $\frac{5}{9}$.(-9) = -5

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

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

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

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

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

Lời giải bài tập Toán 7 Bài 2: Các phép tính với số hữu tỉ hay, chi tiết khác: