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![](https://rs.olm.vn/images/avt/0.png?1311)
`A=(8 2/7-4 2/7)-3 4/9`
`=8+2/7-4-2/7-3-4/9`
`=4-3-4/9`
`=1-4/9=5/9`
`B=(10 2/9-6 2/9)+2 3/5`
`=10+2/9-6-2/9+2+3/5`
`=4+2+3/5`
`=6+3/5=33/5`
Bài 2:
`a)5 1/2*3 1/4`
`=11/2*13/4`
`=143/8`
`b)6 1/3:4 2/9`
`=19/3:38/9`
`=19/3*9/38=3/2`
`c)4 3/7*2`
`=31/7*2`
`=62/7`
Bài 1:
\(A=\left(8\dfrac{2}{7}-4\dfrac{2}{7}\right)-3\dfrac{4}{9}\)
\(A=\left(\dfrac{58}{7}-\dfrac{30}{7}\right)-\dfrac{31}{9}\)
\(A=4-\dfrac{31}{9}\)
\(A=\dfrac{5}{9}\)
\(B=\left(10\dfrac{2}{9}-6\dfrac{2}{9}\right)+2\dfrac{3}{5}\)
\(B=\left(\dfrac{92}{9}-\dfrac{56}{9}\right)+\dfrac{13}{5}\)
\(B=4+\dfrac{13}{5}\)
\(B=\dfrac{33}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
3.a)\(\dfrac{-1}{2}+\dfrac{5}{6}+\dfrac{1}{3}=\dfrac{-3}{6}+\dfrac{5}{6}+\dfrac{2}{6}=\dfrac{-3+5+2}{6}=\dfrac{4}{6}=\dfrac{2}{3}\)
b)\(\dfrac{-3}{8}+\dfrac{7}{4}-\dfrac{1}{12}=\dfrac{-9}{24}+\dfrac{42}{24}-\dfrac{2}{24}=\dfrac{-9+42-2}{24}=\dfrac{31}{24}\)
c)\(\dfrac{3}{5}:\left(\dfrac{1}{4}.\dfrac{7}{5}\right)=\dfrac{3}{5}:\dfrac{7}{20}=\dfrac{3}{5}.\dfrac{20}{7}=\dfrac{12}{7}\)
d)\(\dfrac{10}{11}+\dfrac{4}{11}:4-\dfrac{1}{8}=\dfrac{10}{11}+\dfrac{4}{11}.\dfrac{1}{4}-\dfrac{1}{8}=\dfrac{10}{11}+\dfrac{1}{11}-\dfrac{1}{8}=1-\dfrac{1}{8}=\dfrac{8}{8}-\dfrac{1}{8}=\dfrac{7}{8}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\dfrac{13}{14}\cdot\dfrac{-7}{8}+\dfrac{-3}{2}\)
\(=-\dfrac{13}{16}+\dfrac{-3}{2}\)
\(=-\dfrac{13}{16}+\dfrac{-24}{16}\)
\(=-\dfrac{37}{16}\)
\(b,\dfrac{5}{17}+\dfrac{-15}{34}\cdot\dfrac{2}{5}\)
\(=\dfrac{5}{17}+\dfrac{-3}{17}\)
\(=\dfrac{2}{17}\)
\(c,\dfrac{1}{5}:\dfrac{1}{10}-\dfrac{1}{3}\cdot\left(\dfrac{6}{5}-\dfrac{2}{4}\right)\)
\(=2-\dfrac{1}{3}\cdot\dfrac{7}{10}\)
\(=2-\dfrac{7}{30}\)
\(=\dfrac{53}{30}\)
\(d,\dfrac{-3}{4}:\left(\dfrac{12}{-5}-\dfrac{-7}{10}\right)\)
\(=\dfrac{-3}{4}:\dfrac{-17}{10}\)
\(=\dfrac{15}{34}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Sửa đề : \(M=\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{1\dfrac{1}{6}-\dfrac{7}{6}+\dfrac{7}{10}}\right):\dfrac{2021}{2022}\)
\(M=\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{1\dfrac{1}{6}-\dfrac{7}{6}+\dfrac{7}{10}}\right):\dfrac{2021}{2022}\\ =\left(\dfrac{2\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{7}{11}\right)}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{6}+\dfrac{7}{10}}\right):\dfrac{2021}{2022}\\ =\left(\dfrac{2}{7}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{\dfrac{7}{2}\left(\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}\right)}\right):\dfrac{2021}{2020}\\ =\left(\dfrac{2}{7}-\dfrac{2}{7}\right):\dfrac{2021}{2022}=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Các câu dễ tự làm nha:
\(D=\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
\(D=\dfrac{1}{99}-\dfrac{1}{100}-\dfrac{1}{99}+\dfrac{1}{98}-\dfrac{1}{98}+\dfrac{1}{97}-...-\dfrac{1}{2}+\dfrac{1}{3}-1+\dfrac{1}{2}\)\(D=-\dfrac{1}{100}-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a.\)
\(\dfrac{2}{9}+\dfrac{-3}{10}+-\dfrac{7}{10}=\dfrac{2}{9}-1=\dfrac{2}{9}-\dfrac{9}{9}=-\dfrac{7}{9}\)
\(b.\)
\(\dfrac{-11}{6}+\dfrac{2}{5}+\dfrac{-1}{6}=\left(-\dfrac{11}{6}+-\dfrac{1}{6}\right)+\dfrac{2}{5}=-2+\dfrac{2}{5}=\dfrac{-10}{5}+\dfrac{2}{5}=\dfrac{-8}{5}\)
\(c.\)
\(-\dfrac{5}{8}+\dfrac{12}{7}+\dfrac{13}{8}+\dfrac{2}{7}=\left(-\dfrac{5}{8}+\dfrac{13}{8}\right)+\left(\dfrac{12}{7}+\dfrac{2}{7}\right)=1+2=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=\left(\dfrac{19}{6}-\dfrac{2}{5}\right):\left(\dfrac{29}{6}+\dfrac{7}{10}\right)\)
\(=\dfrac{19\cdot5-2\cdot6}{30}:\dfrac{290+42}{30}=\dfrac{83}{332}=\dfrac{1}{4}\)
b: \(=\dfrac{\left(\dfrac{102}{25}-\dfrac{2}{25}\right)\cdot\dfrac{17}{4}}{\left(6+\dfrac{5}{9}-3-\dfrac{1}{4}\right)\cdot\dfrac{16}{7}}\)
\(=\dfrac{4\cdot\dfrac{17}{4}}{\dfrac{16}{7}\cdot\dfrac{119}{36}}=\dfrac{17}{\dfrac{68}{9}}=17\cdot\dfrac{9}{68}=\dfrac{9}{4}\)
c: \(=\left(\dfrac{120}{60}-\dfrac{15}{60}+\dfrac{20}{60}-\dfrac{36}{60}\right):\left(\dfrac{45}{15}-\dfrac{3}{15}-\dfrac{25}{15}\right)\)
\(=\dfrac{89}{60}:\dfrac{17}{15}=\dfrac{89}{60}\cdot\dfrac{15}{17}=\dfrac{89}{68}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (2x - 3)(6 - 2x) = 0
=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)
b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)
c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)
d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)
e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)
a) \(\dfrac{\dfrac{7}{10}+\dfrac{3}{5}}{\dfrac{7}{10}+\dfrac{1}{2}}\) = (\(\dfrac{7}{10}+\dfrac{3}{5}\) ) : ( \(\dfrac{7}{10}+\dfrac{1}{2}\) )
= \(\dfrac{7+6}{10}\) : \(\dfrac{7+5}{10}\)
= \(\dfrac{13}{10}:\dfrac{12}{10}\)
= \(\dfrac{13}{10}.\dfrac{10}{12}\) =\(\dfrac{13}{12}\)
b) \(\dfrac{6-\dfrac{1}{\dfrac{1}{2}-\dfrac{1}{3}}}{6+\dfrac{1}{\dfrac{1}{2}-\dfrac{1}{3}}}\)
= ( 6 - \(\dfrac{1}{\dfrac{1}{2}-\dfrac{1}{3}}\) ) : ( 6 + \(\dfrac{1}{\dfrac{1}{2}-\dfrac{1}{3}}\) )
= ( 6 - \(\dfrac{1}{\dfrac{1}{6}}\) ) : ( 6 + \(\dfrac{1}{\dfrac{1}{6}}\) )
= ( 6 - 6 ) : (6 + 6) = \(\dfrac{0}{12}\) =0