Sau bài học này chúng ta sẽ giải bài toán trên như sau:

Vì chiều cao tầng hầm B2 bằng $\frac{4}{3}$  tầng hầm B1 nên chiều cao tầng hầm B2 bằng   $2,7\cdot \frac{4}{3}=\frac{27}{10}\cdot \frac{4}{3}=\frac{9.2}{5}=\frac{18}{5}\left(m\right)$

Chiều cao hai tầng hầm của tòa nhà bằng tổng chiều cao tầng B1 với chiều cao tầng B2 và bằng

$2,7+\frac{18}{5}=\frac{27}{10}+\frac{18}{5}=\frac{27}{10}+\frac{36}{10}=\frac{63}{10}=6,3\left(m\right)$

Khi nói về chiều cao của tầng hầm so với mặt đất ta thường dùng số âm để biểu thị sự trái ngược so với chiều cao các tầng của tòa nhà so với mặt đất được biểu thị bởi số âm.

Vậy chiều cao tầng hầm của tòa nhà so với mặt đất là: -6,3m.

Thiết bị lặn ở độ sâu so với mực nước biển bằng:

$\frac{43}{6}+5,4=\frac{43}{6}+\frac{54}{10}=\frac{43}{6}+\frac{27}{5}=\frac{215}{30}+\frac{162}{30}=\frac{377}{30}\left(m\right)$

Ta đã biết khi nói về lặn xuống biển ta sẽ dùng số dương để thể hiện độ sâu và số âm để thể hiện độ cao so với mực nước biển.

Vậy thiết bị lặn đang ở độ cao $-\frac{377}{30}m$ so với mực nước biển .

$$\begin{gathered} a) 0,6+\left(\frac{3}{-4}\right)=\frac{6}{10}+\left(\frac{-3}{4}\right)=\frac{12}{20}+\left(\frac{-15}{20}\right)=\frac{12+(-15)}{20}=\frac{-3}{20} \\ b) \left(-1\frac{1}{3}\right)-(-0,8)=\left(-\frac{1.3+1}{3}\right)-\left(-\frac{8}{10}\right)=\left(-\frac{4}{3}\right)-\left(\frac{-8}{10}\right) \\ =\left(\frac{-4}{3}\right)-\left(\frac{-8}{10}\right)=\left(\frac{-40}{30}\right)-\left(\frac{-24}{30}\right)=\frac{-40-(-24)}{30}=\frac{-16}{30}=\frac{-8}{15} \end{gathered}$$

$$\begin{gathered} a) M=\frac{1}{2}+\frac{2}{3}+\left(-\frac{1}{2}\right)+\frac{1}{3}=\frac{3}{6}+\frac{4}{6}+\left(-\frac{1}{2}\right)+\frac{1}{3}=\frac{7}{6}+\left(-\frac{1}{2}\right)+\frac{1}{3} \\ =\frac{7}{6}+\left(\frac{-3}{6}\right)+\frac{1}{3}=\frac{4}{6}+\frac{1}{3}=\frac{2}{3}+\frac{1}{3}=1 \end{gathered}$$

Vậy M = 1

b) $M=\frac{1}{2}+\frac{2}{3}+\left(-\frac{1}{2}\right)+\frac{1}{3}=\left[\frac{1}{2}+\left(-\frac{1}{2}\right)\right]+\left(\frac{2}{3}+\frac{1}{3}\right)=0+\frac{3}{3}=1$

Vậy M = 1

$$\begin{gathered} B=\left(\frac{-3}{13}\right)+\frac{16}{23}+\left(\frac{-10}{13}\right)+\frac{5}{11}+\frac{7}{23} \\ =\left[\left(\frac{-3}{13}\right)+\left(\frac{-10}{13}\right)\right]+\left[\left(\frac{16}{23}\right)+\left(\frac{7}{23}\right)\right]+\frac{5}{11} \\ =\frac{-13}{13}+\frac{23}{23}+\frac{5}{11}=-1+1+\frac{5}{11}=0+\frac{5}{11}=\frac{5}{11} \end{gathered}$$

Vậy B = $\frac{5}{11}$

Lượng cà phê tồn kho trong 6 tuần là: $32-18,5-5\frac{4}{5}+18,3-12-\frac{39}{4}$

$$\begin{gathered} (32-18,5+18,3-12)+\left[\left(-5\frac{4}{5}\right)+\left(\frac{-39}{4}\right)\right] \\ =19,8+\left[\left(-\frac{5.5+4}{5}\right)+\left(\frac{-39}{4}\right)\right] \\ =19,8+\left(\frac{-29}{5}+\frac{-39}{4}\right) \\ =19,8+\left(\frac{-116}{20}+\frac{-195}{20}\right) \\ =19,8+\frac{-311}{20} \\ =\frac{198}{10}+\frac{-311}{20}=\frac{396}{20}+\frac{-311}{20}=\frac{85}{20}=\frac{17}{4} \left(tấn\right) \end{gathered}$$

Vậy lượng cà phê tồn kho trong 6 tuần là $\frac{17}{4}$ tấn

a) $(-3,5).\left(1\frac{3}{5}\right)=\left(\frac{-35}{10}\right)\cdot \left(\frac{1.5+3}{5}\right)=\frac{-7}{2}\cdot \frac{8}{5}=\frac{-7.8}{2.5}=\frac{-7.4}{5}=\frac{-28}{5}$

b) $\frac{-5}{9}\cdot \left(-2\frac{1}{2}\right)==\frac{-5}{9}\cdot \left(-\frac{2.2+1}{2}\right)=\frac{-5}{9}\cdot \left(-\frac{5}{2}\right)=\frac{-5.(-5)}{9.2}=\frac{25}{18}$

a)

 $$\begin{gathered} M=\frac{1}{7}\cdot \left(\frac{-5}{8}\right)+\frac{1}{7}\cdot \left(\frac{-11}{8}\right) \\ =\frac{1.(-5)}{7.8}+\frac{1.(-11)}{7.8}=\frac{-5}{56}+\frac{-11}{56}=\frac{-16}{56}=\frac{-2}{7} \end{gathered}$$

b)

 $$\begin{gathered} M=\frac{1}{7}\cdot \left(\frac{-5}{8}\right)+\frac{1}{7}\cdot \left(\frac{-11}{8}\right) \\ =\frac{1}{7}.\left[\left(\frac{-5}{8}\right)+\left(\frac{-11}{8}\right)\right]=\frac{1}{7}\cdot \left(\frac{-16}{8}\right)=\frac{1}{7}\cdot (-2)=\frac{-2}{7} \end{gathered}$$

a)

 $$\begin{gathered} A=\frac{5}{11}\cdot \left(\frac{-3}{23}\right)\cdot \frac{11}{5}\cdot (-4,6) \\ =\frac{5}{11}\cdot \left(\frac{-3}{23}\right)\cdot \frac{11}{5}\cdot \left(\frac{-46}{10}\right)=\left(\frac{5}{11}\cdot \frac{11}{5}\right)\cdot \left(\frac{-3}{23}\right)\cdot \left(\frac{-46}{10}\right) \\ =1.\left(\frac{-3}{23}\right)\cdot \left(\frac{-23}{5}\right)=\frac{\left(-3\right).(-23)}{23.5}=\frac{3}{5} \end{gathered}$$

b) 

$$\begin{gathered} B=\left(-\frac{7}{9}\right)\cdot \frac{13}{25}-\frac{13}{25}\cdot \frac{2}{9} \\ =\frac{13}{25}\cdot \left[\left(-\frac{7}{9}\right)-\frac{2}{9}\right]=\frac{13}{25}\cdot (-1)=\frac{-13}{25} \end{gathered}$$

Vì chiều cao tầng hầm B2 bằng $\frac{4}{3}$  tầng hầm B1 nên chiều cao tầng hầm B2 bằng $2,7\cdot \frac{4}{3}=\frac{27}{10}\cdot \frac{4}{3}=\frac{9.2}{5}=\frac{18}{5}\left(m\right)$

Chiều cao tầng hầm của tòa nhà bằng tổng chiều cao tầng B1 với chiều cao tầng B2 và bằng

$2,7+\frac{18}{5}=\frac{27}{10}+\frac{18}{5}=\frac{27}{10}+\frac{36}{10}=\frac{63}{10}=6,3\left(m\right)$.

Khi nói về chiều cao của tầng hầm so với mặt đất ta thường dùng số âm để biểu thị sự trái ngược so với chiều cao các tầng của tòa nhà so với mặt đất được biểu thị bở số âm.

Vậy chiều cao tầng hầm của tòa nhà so với mặt đất là: -6,3m.

a) $\frac{14}{15}:\left(-\frac{7}{5}\right)=\frac{14}{15}.\left(\frac{-5}{7}\right)=\frac{14.(-5)}{15.7}=\frac{-2}{3}$

b) $\left(-2\frac{2}{5}\right):(-0,32)=\left(\frac{-12}{5}\right):\left(\frac{-32}{100}\right)$

$=\left(\frac{-12}{5}\right):\left(\frac{-8}{25}\right)=\frac{(-12).25}{5.(-8)}=\frac{3.5}{2}=\frac{15}{2}$

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

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

$c) \left(\frac{-7}{12}\right)+0,75=\left(\frac{-7}{12}\right)+\frac{75}{100}=\left(\frac{-7}{12}\right)+\frac{9}{12}=\frac{2}{12}=\frac{1}{6}$

$d) \left(\frac{-5}{9}\right)-1,25=\left(\frac{-5}{9}\right)-\frac{125}{100}=\left(\frac{-5}{9}\right)-\frac{5}{4}=\left(\frac{-20}{36}\right)-\frac{45}{36}=\frac{-65}{36}$

$e) 0,34\cdot \left(\frac{-5}{17}\right)=\frac{34}{100}\cdot \left(\frac{-5}{17}\right)=\frac{17}{50}\cdot \left(\frac{-5}{17}\right)=\frac{-5}{50}=\frac{-1}{10}$

$g) \frac{4}{9}:\left(\frac{-8}{15}\right)=\frac{4}{9}\cdot \left(\frac{-15}{8}\right)=\frac{4.\left(-15\right)}{9.8}=\frac{-5}{3.2}=\frac{-5}{6}$

$h) \left(1\frac{2}{3}\right):\left(2\frac{1}{2}\right)=\left(\frac{1.3+2}{3}\right):\left(\frac{2.2+1}{2}\right)=\frac{5}{3}:\frac{5}{2}=\frac{5}{3}\cdot \frac{2}{5}=\frac{2}{3}$

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

$k) \left(\frac{-3}{5}\right)\cdot \left(\frac{15}{-7}\right)\cdot 3\frac{1}{9}=\left(\frac{-3}{5}\right)\cdot \left(\frac{15}{-7}\right)\cdot \frac{28}{9}=\frac{(-3)\mathrm{.15.28}}{5.(-7).9}=\frac{(-3)\mathrm{.3.5.7.4}}{5.(-7).9}=4$

a) $0,75-\frac{5}{6}+1\frac{1}{2}=\frac{75}{100}-\frac{5}{6}+\frac{3}{2}=\frac{3}{4}-\frac{5}{6}+\frac{3}{2}$

$$\begin{gathered} b) \frac{3}{7}+\frac{4}{15}+\left(\frac{-8}{21}\right)+(-0,4)=\frac{3}{7}+\left(\frac{-8}{21}\right)+\frac{4}{15}+\frac{-4}{10} \\ =\left(\frac{9}{21}+\frac{-8}{21}\right)+\frac{4}{15}+\frac{-2}{5}=\frac{1}{21}+\frac{4}{15}+\frac{-6}{15}=\frac{1}{21}+\frac{-2}{15}=\frac{5}{105}+\frac{-14}{105}=\frac{-9}{105}=\frac{-3}{35} \end{gathered}$$

$$\begin{gathered} c) 0,625+\left(\frac{-2}{7}\right)+\frac{3}{8}+\left(\frac{-5}{7}\right)+1\frac{2}{3}=\frac{625}{1000}+\left(\frac{-2}{7}\right)+\frac{3}{8}+\left(\frac{-5}{7}\right)+\frac{5}{3} \\ =\frac{5}{8}+\left(\frac{-2}{7}\right)+\frac{3}{8}+\left(\frac{-5}{7}\right)+\frac{5}{3}=\left(\frac{5}{8}+\frac{3}{8}\right)+\left[\left(\frac{-2}{7}\right)+\left(\frac{-5}{7}\right)\right]+\frac{5}{3} \\ =1+(-1)+\frac{5}{3}=\frac{5}{3} \end{gathered}$$

$$\begin{gathered} d) (-3)\cdot \left(\frac{-38}{21}\right)\cdot \left(\frac{-7}{6}\right)\cdot \left(-\frac{3}{19}\right)=\frac{(-3).(-38).(-7).(-3)}{\mathrm{21.6.19}} \\ =\frac{(-3).(-7).(-38).(-3)}{\mathrm{21.6.19}}=\frac{\mathrm{3.7.2.19.3}}{\mathrm{3.7.2.3.19}}=1 \end{gathered}$$

$e) \left(\frac{11}{18}:\frac{22}{9}\right)\cdot \frac{8}{5}=\frac{11}{18}\cdot \frac{9}{22}\cdot \frac{8}{5}=\frac{\mathrm{11.9.8}}{\mathrm{9.2.11.2.5}}=\frac{2}{5}$

$g) \left[\left(\frac{-4}{5}\right)\cdot \frac{5}{8}\right]:\left(\frac{-25}{12}\right)=\frac{\left(-4\right).5}{5.8}:\left(\frac{-25}{12}\right)=\frac{-1}{2}:\left(\frac{-25}{12}\right)=\frac{-1}{2}.\frac{12}{-25}=\frac{6}{25}$

 $$\begin{gathered} a) \left(\frac{-5}{8}\right)+\left(\frac{3}{-8}\right)=\left(\frac{-5}{8}\right)+\left(\frac{-3}{8}\right)=\frac{-8}{8}=-1 \\ Vậy ta điền \left(\frac{-5}{8}\right)+\left(\frac{3}{-8}\right)\boxed{=}-1 \end{gathered}$$

$$\begin{gathered} b) \left(\frac{-13}{22}\right)+\left(\frac{-5}{22}\right)=\frac{-18}{22}=\frac{-9}{11}<\frac{-8}{11} \\ Vậy ta điền \left(\frac{-13}{22}\right)+\left(\frac{-5}{22}\right)\boxed{<}\frac{-8}{11} \end{gathered}$$

$$\begin{gathered} c) \frac{1}{6}+\left(\frac{-3}{4}\right)=\frac{2}{12}+\left(\frac{-9}{12}\right)=\frac{-7}{12} \\ \frac{1}{14}+\left(\frac{-4}{7}\right)=\frac{1}{14}+\left(\frac{-8}{14}\right)=\frac{-7}{14} \\ Ta có \frac{-7}{12}<\frac{-7}{14} \\ Vậy ta điền \frac{1}{6}+\left(\frac{-3}{4}\right)\boxed{<} \frac{1}{14}+\left(\frac{-4}{7}\right). \end{gathered}$$

$$\begin{gathered} a) \frac{3}{7}\cdot \left(\frac{-1}{9}\right)+\frac{3}{7}\cdot \left(-\frac{2}{3}\right) \\ =\frac{3}{7}\cdot \left[\left(\frac{-1}{9}\right)+\left(-\frac{2}{3}\right)\right] \\ =\frac{3}{7}\cdot \left[\left(\frac{-1}{9}\right)+\left(\frac{-6}{9}\right)\right] \\ =\frac{3}{7}\cdot \left(\frac{-7}{9}\right)=\frac{-1}{3} \end{gathered}$$

$$\begin{gathered} b) \left(\frac{-7}{13}\right)\cdot \frac{5}{12}+\left(\frac{-7}{13}\right)\cdot \frac{7}{12}+\left(\frac{-6}{13}\right) \\ =\left(\frac{-7}{13}\right)\cdot \left(\frac{5}{12}+\frac{7}{12}\right)+\left(\frac{-6}{13}\right) \\ =\left(\frac{-7}{13}\right)\cdot \frac{12}{12}+\left(\frac{-6}{13}\right) \\ =\left(\frac{-7}{13}\right)+\left(\frac{-6}{13}\right)=\frac{-13}{13}=-1 \end{gathered}$$

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

$$\begin{gathered} d) \frac{5}{9}:\left(\frac{1}{11}-\frac{5}{22}\right)+\frac{5}{9}:\left(\frac{1}{15}-\frac{2}{3}\right)=\frac{5}{9}:\left(\frac{2}{22}-\frac{5}{22}\right)+\frac{5}{9}:\left(\frac{1}{15}-\frac{10}{15}\right) \\ =\frac{5}{9}:\left(\frac{-3}{22}\right)+\frac{5}{9}:\left(\frac{-9}{15}\right)=\frac{5}{9}\cdot \frac{-22}{3}+\frac{5}{9}\cdot \frac{-15}{9}=\frac{5}{9}\cdot \left(\frac{-22}{3}+\frac{-5}{3}\right)=\frac{5}{9}\cdot (-9)=-5 \end{gathered}$$

$$\begin{gathered} e) \frac{3}{5}+\frac{3}{11}-\left(\frac{-3}{7}\right)+\left(\frac{-2}{97}\right)-\frac{1}{35}-\frac{3}{4}+\left(\frac{-23}{44}\right) \\ =\left[\frac{3}{5}-\left(\frac{-3}{7}\right)-\frac{1}{35}\right]+\left[\frac{3}{11}-\frac{3}{4}+\left(\frac{-23}{44}\right)\right]+\left(\frac{-2}{97}\right) \\ =\left[\frac{21}{35}-\left(\frac{-15}{35}\right)-\frac{1}{35}\right]+\left[\frac{12}{44}-\frac{33}{44}+\left(\frac{-23}{44}\right)\right]+\left(\frac{-2}{97}\right) \\ =1+(-1)+\left(\frac{-2}{97}\right)=\frac{-2}{97} \end{gathered}$$

$$\begin{gathered} a) x\cdot \frac{14}{27}=\frac{-7}{9}; \\ x=\frac{-7}{9}:\frac{14}{27} \\ x=\frac{-7}{9}\cdot \frac{27}{14} \\ x=\frac{-3}{2} \\ Vậy x=\frac{-3}{2} \end{gathered}$$

$$\begin{gathered} b) \left(\frac{-5}{9}\right):x=\frac{2}{3} \\ x=\left(-\frac{5}{9}\right):\frac{2}{3} \\ x=\left(-\frac{5}{9}\right)\cdot \frac{3}{2} \\ x=\frac{-5}{6} \\ Vậy x=\frac{-5}{6} \end{gathered}$$

$$\begin{gathered} c) \frac{2}{5}:x=\frac{1}{16}:0,125 \\ \frac{2}{5}:x=\frac{1}{16}:\frac{125}{1000} \\ \frac{2}{5}:x=\frac{1}{16}:\frac{1}{8} \\ \frac{2}{5}:x=\frac{1}{16}\cdot 8 \\ x=\frac{2}{5}:\frac{1}{2} \\ x=\frac{2}{5}\cdot 2 \\ x=\frac{4}{5} \\ Vậy x=\frac{4}{5} \end{gathered}$$

$$\begin{gathered} d) -\frac{5}{12}x=\frac{2}{3}-\frac{1}{2}. \\ \frac{-5}{12}x=\frac{4}{6}-\frac{3}{6} \\ \frac{{\displaystyle -5}}{{\displaystyle 12}}x=\frac{{\displaystyle 1}}{{\displaystyle 6}} \\ x=\frac{{\displaystyle 1}}{{\displaystyle 6}}:\frac{{\displaystyle -5}}{{\displaystyle 12}} \\ x=\frac{{\displaystyle 1}}{{\displaystyle 6}}\cdot \frac{{\displaystyle 12}}{{\displaystyle -5}} \\ x=\frac{{\displaystyle -2}}{{\displaystyle 5}} \\ Vậy x=\frac{{\displaystyle -2}}{{\displaystyle 5}} \end{gathered}$$