Bài tập: Nhân, Chia số hữu tỉ

$\frac{-25}{16}=\frac{-5}{12}.\frac{15}{4}$

$\frac{-25}{16}=\frac{-4}{5}:\frac{64}{125}.$

$\left(-\frac{28}{19}\right).\left(\frac{-38}{14}\right)=\frac{(-28).(-38)}{19.14}=2.2=4$

$$\begin{gathered} \left(-\frac{-21}{16}\right).\left(-\frac{24}{7}\right)=\frac{21.(-24)}{16.7} \\ =\frac{3.(-3)}{2}=\frac{-9}{2}=-4\frac{1}{2} \end{gathered}$$

$$\begin{gathered} \frac{-12}{7}.1,25=\frac{-12}{7}.\frac{125}{100}=\frac{-12}{7}.\frac{5}{4} \\ =\frac{-12.5}{7.4}=\frac{-3.5}{7}=\frac{-15}{7}=-2\frac{1}{7} \end{gathered}$$

$\frac{-4}{5}\cdot 3\frac{2}{16}=\frac{-4}{5}.\frac{50}{16}=\frac{-4.50}{5.16}=\frac{-10}{4}=\frac{-5}{2}=-2\frac{1}{2}$

$12.{\left(-\frac{2}{3}\right)}^2=12.\frac{4}{9}=\frac{12.4}{9}=\frac{4.4}{3}=\frac{16}{3}=5\frac{1}{3}$

$\begin{array}{l}(-2)\cdot \frac{-38}{21}\cdot \frac{-7}{4}\cdot \left(-\frac{3}{8}\right)=\frac{(-2).(-38).(-7).(-3)}{\mathrm{21.4.8}}\\ =\frac{\mathrm{2.38.7.3}}{\mathrm{21.4.8}}=\frac{19}{8}=2\frac{3}{8}\end{array}$

$2\frac{2}{3}\cdot \frac{2}{5}\cdot \frac{3}{8}\cdot 10\cdot \frac{19}{92}=\frac{8}{3}.\frac{2}{5}.\frac{3}{8}\mathrm{.10.}\frac{19}{92}=\frac{\mathrm{8.2.3.10.19}}{\mathrm{3.5.8.92}}=\frac{19}{23}$

$P=\frac{-8}{13}\cdot \frac{16}{9}\cdot \frac{-13}{8}\cdot \left(-18\right)=\frac{(-8)\mathrm{.16.}(-13).(-18)}{\mathrm{13.9.8}}=-32$

$\frac{-4}{3}:\frac{13}{9}=\frac{-4}{3}.\frac{9}{13}=\frac{-4.9}{3.13}=\frac{-12}{13}$

$\left(-\frac{9}{25}\right):6=\frac{-9}{25}.\frac{1}{6}=\frac{(-9).1}{25.6}=\frac{-3}{50}$

$\frac{11}{12}:2\frac{1}{16}=\frac{11}{12}:\frac{33}{16}=\frac{11}{12}.\frac{16}{33}=\frac{11.16}{12.33}=\frac{4}{9}$

$3,5:\frac{-3}{2}=\frac{35}{10}:\frac{-3}{2}=\frac{7}{2}.\frac{-2}{3}=\frac{7.(-2)}{2.3}=\frac{-7}{3}=-2\frac{1}{3}$

$\begin{array}{l}\frac{-1}{32}:(-4):2\frac{1}{3}=\frac{-1}{32}:(-4):\frac{7}{3}\\ =\frac{-1}{32}.\frac{-1}{4}.\frac{3}{7}=\frac{\mathrm{1.1.3}}{\mathrm{32.4.7}}=\frac{3}{896}\end{array}$

$\frac{-3}{4}\cdot \frac{12}{-5}:\left(-\frac{25}{16}\right)=\frac{-3}{4}\cdot \frac{12}{-5}\cdot \frac{-16}{25}=\frac{9}{5}.\frac{-16}{25}=\frac{-144}{125}$

$\frac{-5}{7}\cdot \frac{49}{3}:\frac{-7}{6}=\frac{-5}{7}\cdot \frac{49}{3}\cdot \frac{-6}{7}=10$

$1,75:\left(-12\right)\cdot \left(\frac{-2}{3}\right)=\frac{7}{4}.\frac{-1}{12}.\frac{-2}{3}=\frac{7}{72}$

$\left(-9\right)\cdot \left(\frac{2}{3}\cdot \frac{5}{4}\right):\left(-7\right)=-9.\frac{5}{6}.\frac{-1}{7}=\frac{15}{14}$

$0,8:\left(-2\frac{4}{5}\right)\cdot {\left(\frac{-5}{2}\right)}^2=\frac{4}{5}:\frac{-14}{5}.\frac{25}{4}=\frac{4}{5}.\frac{-5}{14}.\frac{25}{4}=\frac{-25}{14}$

$21-3\frac{3}{4}:\left(\frac{3}{8}-\frac{1}{6}\right)=21-\frac{15}{4}:\frac{5}{24}=21-18=3$

$\frac{7}{23}\cdot \left[\left(-\frac{8}{6}\right)-\frac{45}{18}\right]=\frac{7}{23}.\frac{-23}{6}=\frac{-7}{6}$

$\left(0,75-\frac{1}{4}\right):\frac{-5}{6}=\left(\frac{3}{4}-\frac{1}{4}\right).\frac{-6}{5}=\frac{1}{2}.\frac{-6}{5}=\frac{-3}{5}$

$\begin{array}{l}\frac{5}{4}-\frac{5}{4}:\left(\frac{3}{8}-\frac{1}{6}\right)+\frac{11}{12}\\ =\frac{5}{4}-\frac{5}{4}:\frac{5}{24}+\frac{11}{12}=\frac{5}{4}-6+\frac{11}{12}\\ =\frac{-19}{4}+\frac{11}{12}=\frac{-23}{6}\end{array}$

$1,25\cdot \left(5-\frac{4}{3}\right)\cdot \left(\frac{-7}{11}\right)=\frac{5}{4}.\frac{11}{3}.\frac{-7}{11}=\frac{-35}{12}$

$\left(-9\right)\cdot \left(\frac{2}{3}-\frac{5}{4}\right):\left(-7\right)=-9.\frac{-7}{12}.\frac{-1}{7}=\frac{-3}{4}$

$\left(1\frac{3}{4}-0,25\right).2\frac{1}{3}=\left(\frac{7}{4}-\frac{1}{4}\right).\frac{7}{3}=\frac{3}{2}.\frac{7}{3}=\frac{7}{2}$

$\left(2\frac{3}{4}-0,25\right).2\frac{1}{3}=\left(\frac{11}{4}-\frac{1}{4}\right).\frac{7}{3}=\frac{5}{2}.\frac{7}{3}=\frac{35}{6}$

$\left(\frac{7}{3}-\frac{5}{2}\right):\left(\frac{3}{4}+\frac{3}{2}\right)=\frac{-1}{6}:\frac{9}{4}=\frac{-1}{6}.\frac{4}{9}=\frac{-2}{27}$

$\begin{array}{l}\left(\frac{9}{25}-2.18\right):\left(3\frac{4}{5}+0,2\right)=\left(\frac{9}{25}-36\right):\left(\frac{19}{5}+\frac{1}{5}\right)\\ =\frac{-891}{25}:4=\frac{-891}{100}\end{array}$

$\begin{array}{l}\frac{0,84-0,64+\frac{4}{9}+\frac{4}{13}}{\frac{7}{9}+\frac{7}{13}-0,2+0,55}=\frac{0,2+\frac{4}{9}+\frac{4}{13}}{\frac{7}{9}+\frac{7}{13}+0,35}\\ =\frac{\frac{4}{20}+\frac{4}{9}+\frac{4}{13}}{\frac{7}{9}+\frac{7}{13}+\frac{7}{20}}\\ =\frac{4\left(\frac{1}{20}+\frac{1}{9}+\frac{1}{13}\right)}{7\left(\frac{1}{20}+\frac{1}{9}+\frac{1}{13}\right)}=\frac{4}{7}\end{array}$

$\begin{array}{l} \frac{3}{8}.19\frac{1}{3}-\frac{3}{8}.33\frac{1}{3} =\frac{3}{8}.\left(19\frac{1}{3}-33\frac{1}{3} \right)\\ =\frac{3}{8}.\left(-14\right)=\frac{-21}{4}\end{array}$

$\begin{array}{l}12,5.\left(-\frac{5}{7}\right)+1,5.\left(-\frac{5}{7}\right)\\ =-\frac{5}{7}.\left(12,5+1,5\right)=-\frac{5}{7}.14=-10\end{array}$

$\begin{array}{l}\frac{3}{5}:\left(-1\frac{1}{15}-\frac{1}{6}\right)+\frac{3}{5}:\left(\frac{-1}{3}-1\frac{1}{15}\right)\\ =\frac{3}{5}:\left(\frac{-16}{15}-\frac{1}{6}\right)+\frac{3}{5}:\left(\frac{-1}{3}-\frac{16}{15}\right)\\ =\frac{3}{5}:\left(-\frac{37}{30}\right)+\frac{3}{5}:\frac{-7}{5}\\ =\frac{3}{5}.\frac{-30}{37}+\frac{3}{5}.\frac{-5}{7}=\frac{3}{5}.\left(\frac{-30}{37}+\frac{-5}{7}\right)\\ =\frac{3}{5}.\frac{-395}{259}\\ =\frac{-237}{259}\end{array}$

$\begin{array}{l}\left(\frac{-3}{4}+\frac{2}{5}\right):\frac{3}{7}+\left(\frac{3}{5}+\frac{-1}{4}\right):\frac{3}{7}\\ =\left(\frac{-3}{4}+\frac{2}{5}\right).\frac{7}{3}+\left(\frac{3}{5}+\frac{-1}{4}\right).\frac{7}{3}\\ =\frac{7}{3}.\left(\frac{-3}{4}+\frac{2}{5}+\frac{3}{5}+\frac{-1}{4}\right)\\ =\frac{7}{3}.\left(\frac{-3}{4}+\frac{-1}{4}+\frac{2}{5}+\frac{3}{5}\right)\\ =\frac{7}{3}.\left(-1+1\right)\\ =0\end{array}$

$\begin{array}{l}A=\left[0,25\times 7+{\left(0,25\right)}^2\right]\left(0,4\times 5^2-\frac{2}{5}\right)\\ A=\left(\frac{7}{4}+\frac{1}{16}\right).\left(\frac{2}{5}.25-\frac{2}{5}\right)\\ A=\frac{29}{16}.\frac{48}{5}\\ A=\frac{87}{5}=17\frac{2}{5}\end{array}$

$\begin{array}{l}B=\left(\frac{2}{89}-\frac{3}{178}\right)\cdot \frac{89}{17}+\frac{33}{34}\\ B=\left(\frac{4}{178}-\frac{3}{178}\right).\frac{89}{17}+\frac{33}{34}\\ B=\frac{1}{178}.\frac{89}{17}+\frac{33}{34}\\ B=\frac{1}{34}+\frac{33}{34}=1\end{array}$

Vậy A>B

$\begin{array}{l}A=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)\mathrm{...}\left(1-\frac{1}{19}\right).\left(1-\frac{1}{20}\right)\\ A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}\mathrm{....}\frac{18}{19}.\frac{19}{20}\\ A=\frac{1}{20}>\frac{1}{21}\end{array}$

Vậy A>$\frac{1}{21}$

$\begin{array}{l}B=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\mathrm{...}.\frac{99}{100}\\ = \frac{1}{100}<1\end{array}$

Vậy B < 1

$\frac{5}{2}x=\frac{-3}{10}-\frac{-4}{5}=>\frac{5}{2}x=\frac{1}{2}=>x=\frac{1}{2}:\frac{5}{2}=>x=\frac{1}{5}$

$\frac{5}{8}:x=\frac{1}{12}-\frac{4}{3}\Rightarrow \frac{5}{8}:x=\frac{-5}{4}=>x=\frac{5}{8}:\frac{-5}{4}=\frac{-1}{2}$

Từ đề bài ta có $x-\frac{1}{3}=0$ hoặc x+$\frac{2}{5}$=0. Tìm được $\mathrm{x}=\frac{1}{3}$hoặc x=$-\frac{2}{5}$

x = 3/4 hoặc x = 2/5

$\begin{array}{l}A=\left(1-\frac{1}{1.2}\right)+\left(1-\frac{1}{2.3}\right)+\mathrm{...}+\left(1-\frac{1}{2015.2016}\right)\\ =1+1+1+\mathrm{...1}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\mathrm{...}+\frac{1}{2015.2016}\right)\\ =2015-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\mathrm{...}+\frac{1}{2015}-\frac{1}{2016}\right)\\ =2015-\left(1-\frac{1}{2016}\right)\\ =2015-1+\frac{1}{2016}\\ =2014+\frac{1}{2016}\\ =\frac{4060225}{2016}\end{array}$

$\begin{array}{l}B=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+\frac{9}{16.25}+\frac{11}{25.36}\\ =1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+\frac{1}{16}-\frac{1}{25}+\frac{1}{25}-\frac{1}{36}\\ =1-\frac{1}{36}\\ =\frac{35}{36}\end{array}$

$\begin{array}{l}C=\frac{3}{1.4}+\frac{3}{4.7}+\mathrm{...}+\frac{3}{100.103}\\ =1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\mathrm{...}+\frac{1}{100}-\frac{1}{103}\\ =1-\frac{1}{103}\\ =\frac{102}{103}\end{array}$

$\begin{array}{l}D=\frac{3}{1.4}+\frac{6}{4.10}+\frac{9}{10.19}+\frac{12}{19.31}+\frac{15}{31.46}+\frac{18}{46.64}\\ =\frac{4-1}{1.4}+\frac{10-4}{4.10}+\frac{19-10}{10.19}+\frac{31-19}{19.31}+\frac{46-31}{31.46}+\frac{64-46}{46.64}\\ =1-\frac{1}{4}+\frac{1}{4}-\frac{1}{10}+\frac{1}{10}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}+\frac{1}{31}-\frac{1}{46}+\frac{1}{46}-\frac{1}{64}\\ =1-\frac{1}{64}\\ =\frac{63}{64}\end{array}$

Ta có:

$\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+\mathrm{...}+\frac{1}{49.50}$

$\begin{array}{l}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots \cdots +\frac{1}{49}-\frac{1}{50}\\ =\left(1+\frac{1}{3}+\frac{1}{5}+\cdots \cdots +\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots \cdots +\frac{1}{50}\right)\\ =\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\cdots \cdots +\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots \cdots +\frac{1}{50}\right)\\ =\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\cdots \cdots +\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\cdots \cdots +\frac{1}{25}\right)\\ =\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+\mathrm{...}+\frac{1}{50}\left(dpcm\right)\end{array}$