Bài tập cuối tuần Toán 7 Tuần 3 có đáp án

a) $-6.\left(-\frac{2}{3}\right).0,25=4.\frac{1}{4}=1$

b) $-\frac{15}{4}.\left(\frac{-7}{15}\right).\left(-2\frac{2}{5}\right)=\frac{7}{4}.\left(-\frac{12}{5}\right)=-\frac{21}{5}$

a) 

$\begin{array}{l}\left(-2\frac{1}{5}\right).\left(-\frac{9}{11}\right).\left(-1\frac{1}{14}\right).\frac{2}{5}\\ =\left(-\frac{11}{5}\right).\left(-\frac{9}{11}\right).\left(-\frac{15}{14}\right).\frac{2}{5}\\ =-\frac{27}{35}\end{array}$

b)

$\begin{array}{l}\left(-5\frac{1}{2}\right).\left(-\frac{1}{2}\right)-\frac{2}{3}.\left(-\frac{2}{3}\right)\\ =\left(-\frac{11}{2}\right).\left(-\frac{1}{2}\right)+\frac{4}{9}=\frac{11}{4}+\frac{4}{9}\\ =\frac{115}{36}\end{array}$

a) 

$\begin{array}{l}\left(1\frac{1}{4}\right).\left(-\frac{8}{15}\right)-\frac{3}{5}+\frac{2}{5}.\left(-\frac{3}{4}\right)\\ =\left(\frac{5}{4}\right).\left(-\frac{8}{15}\right)-\frac{3}{5}-\frac{3}{10}\\ =\left(-\frac{2}{3}\right)-\frac{3}{5}-\frac{3}{10}\\ =-\frac{47}{30}\end{array}$

b)

$\begin{array}{l}\left(-0,125\right).(-16).\left(-\frac{8}{9}\right).\left(-0,25\right)\\ =\frac{1}{8}\mathrm{.16.}\frac{8}{9}.\frac{1}{4}=\frac{4}{9}\end{array}$

a) 

$\begin{array}{l}\frac{5}{8}+2\frac{1}{4}.1\frac{2}{3}-\frac{1}{4}.\frac{5}{6}\\ =\frac{5}{8}+\frac{9}{4}.\frac{5}{3}-\frac{5}{24}\\ =\frac{5}{4}.\left(\frac{1}{2}+3-\frac{1}{6}\right)\\ =\frac{5}{4}.\frac{10}{3}=\frac{25}{6}\end{array}$

b)

$\begin{array}{l}\left(13\frac{9}{11}:\frac{38}{49}-5\frac{2}{11}:\frac{38}{49}\right):\left(\frac{49}{38}.\frac{5}{11}\right)\\ =\left\{(13+\frac{9}{11}).\frac{49}{38}-(5+\frac{2}{11}).\frac{49}{38}\right\}:\left(\frac{49}{38}.\frac{5}{11}\right)\\ =\frac{49}{38}.\left(13-5+\frac{9}{11}-\frac{2}{11}\right):\left(\frac{49}{38}.\frac{5}{11}\right)\\ =\frac{49}{38}.\left(8+\frac{7}{11}\right):\left(\frac{49}{38}.\frac{5}{11}\right)\\ =\left(8+\frac{7}{11}\right):\frac{5}{11}=19\end{array}$

a) 

$\begin{array}{l}\frac{11}{30}+\frac{18}{35}.\left(\frac{35}{54}-\frac{49}{18}-\frac{28}{48}\right)\\ =\frac{11}{30}+\left(\frac{18}{35}.\frac{35}{54}-\frac{18}{35}.\frac{49}{18}-\frac{18}{35}.\frac{28}{48}\right)\\ =\frac{11}{30}+\left(\frac{1}{3}-\frac{7}{5}-\frac{3}{10}\right)\\ =\frac{11}{30}+\frac{-41}{30}=-1\end{array}$

b) $\frac{-23}{39}.\frac{-13}{56}.\frac{70}{23}:\frac{125}{75}=\frac{5}{3.4}.\frac{3}{5}=\frac{1}{4}$

$\begin{array}{l}a)\frac{-1}{10}+\frac{2}{5}x+\frac{7}{20}=\frac{1}{10}\\ \frac{2}{5}x=\frac{1}{10}-\frac{7}{20}+\frac{1}{10}\\ \frac{2}{5}x=-\frac{3}{20}\\ x=-\frac{3}{8}\end{array}$

$\begin{array}{l}b)\frac{1}{3}+\frac{1}{2}:x=-\frac{1}{5}\\ \frac{1}{2}:x=-\frac{1}{5}-\frac{1}{3}\\ \frac{1}{2}:x=-\frac{8}{15}\\ x=-\frac{15}{16}\end{array}$

$\begin{array}{l}a)-\frac{2}{3}:x+\frac{5}{8}=-\frac{7}{12}\\ \frac{-2}{3}:x=-\frac{7}{12}-\frac{5}{8}\\ \frac{-2}{3}:x=-\frac{29}{24}\\ x=\frac{16}{29}\end{array}$

$\begin{array}{l}b)\frac{1}{2}x+2\frac{1}{2}=3\frac{1}{2}x-\frac{3}{4}\\ \left(\frac{1}{2}-3\frac{1}{2}\right)x=-\frac{3}{4}-\frac{5}{4}\\ -3x=-2\\ x=\frac{2}{3}\end{array}$

$\begin{array}{l}a)\frac{2}{3}x-\frac{2}{5}=\frac{1}{2}x-\frac{1}{3}\\ \left(\frac{2}{3}-\frac{1}{2}\right)x=-\frac{1}{3}+\frac{2}{5}\\ \frac{1}{6}x=\frac{1}{15}\\ x=\frac{2}{5}\end{array}$

$\begin{array}{l}b)\frac{1}{3}x+\frac{2}{5}\left(x+1\right)=0\\ \left(\frac{1}{3}+\frac{2}{5}\right)x=-\frac{2}{5}\\ \frac{11}{15}x=-\frac{2}{5}\\ x=-\frac{6}{11}\end{array}$

$$\begin{gathered} \left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)\mathrm{.....}\left(\frac{1}{1999}-1\right) \\ =\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}\mathrm{....}\frac{-1998}{1999} \\ =\frac{1}{1999} \end{gathered}$$

$$\begin{gathered} \frac{5.18-10.27+15.36}{10.36-20.54+30.72} \\ =\frac{5.18-\mathrm{5.18.3}+\mathrm{5.18.6}}{10.36-\mathrm{10.36.3}+\mathrm{10.36.6}} \\ =\frac{5.18(1-3+6)}{10.36(1-3+6)} \\ =\frac{1}{4} \end{gathered}$$

$$\begin{gathered} -1\frac{1}{2}.\left(-1\frac{1}{3}\right).\left(-1\frac{1}{4}\right)\mathrm{......}\left(-1\frac{1}{1999}\right) \\ =\frac{-3}{2}.\frac{-4}{3}.\frac{-5}{4}\mathrm{.....}\frac{-2000}{1999} \\ =1000 \end{gathered}$$

Xét các góc tạo bởi đường thẳng a và cát tuyến c

Nhận xét: Theo hình vẽ trên ta có:

Hai góc so le trong bất kỳ bằng nhau.

Hai góc đồng bị bất kỳ bằng nhau.

Hai góc trong cùng phía bù nhau. (Tổng hai góc trong cùng phía bằng ${180}^o$)