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a: \(\dfrac{-13}{40}< \dfrac{-12}{40}\)
\(\dfrac{-5}{6}>\dfrac{-91}{104}\)
a) 15/11 - (5/7 - 18/11) + 27/7
= 15/11 - 5/7 + 18/11 + 27/7
= (15/11 + 18/11) + (-5/7 + 27/7)
= 3 + 22/7
= 43/7
b) 39/5 + (9/4 - 9/5) - (5/4 + 1,2)
= 39/5 + 9/4 - 9/5 - 5/4 - 6/5
= (39/5 - 9/5 - 6/5) + (9/4 - 5/4)
= 24/5 + 1
= 29/5
c) -1,2 - 0,8 + 0,25 + 5,75 - 2022
= (-1,2 - 0,8) + (0,25 + 5,76) - 2022
= -2 + 6 - 2022
= 4 - 2022
= -2018
d) 0,1 + 16/9 + 5,1 + (-20/9)
= (0,1 + 5,1) + (16/9 - 20/9)
= 5,2 - 4/9
= 419/90
a) \(\dfrac{15}{11}-\left(\dfrac{5}{7}-\dfrac{18}{11}\right)+\dfrac{27}{7}=\dfrac{22}{7}+3=\dfrac{43}{77}\)
b) \(\dfrac{39}{5}+\left(\dfrac{9}{4}-\dfrac{9}{5}\right)-\left(\dfrac{5}{4}+\dfrac{6}{5}\right)=\dfrac{24}{5}+1=\dfrac{29}{5}\)
c) \(-1,2-0,8+0,25+5,75-2022=-2+6-2022=-2018\)
d) \(0,1+\dfrac{16}{9}+5,1+\dfrac{-20}{9}=\dfrac{26}{5}-\dfrac{4}{9}=\dfrac{214}{45}\)
a) Ta có:
2A=2.(12+122+123+...+122020+122021)2�=2.12+122+123+...+122 020+122 021
2A=1+12+122+123+...+122019+1220202�=1+12+122+123+...+122 019+122 020
Suy ra: 2A−A=(1+12+122+123+...+122019+122020)2�−�=1+12+122+123+...+122 019+122 020
−(12+122+123+...+122020+122021)−12+122+123+...+122 020+122 021
Do đó A=1−122021<1�=1−122021<1.
Lại có B=13+14+15+1360=20+15+12+1360=6060=1�=13+14+15+1360=20+15+12+1360=6060=1.
Vậy A < B.
`@` `\text {Ans}`
`\downarrow`
`a)`
`-9/34*17/4`
`=`\(\dfrac{-9}{17\cdot2}\cdot\dfrac{17}{4}\)
`=`\(-\dfrac{9}{2}\cdot\dfrac{1}{4}\)
`=`\(-\dfrac{9}{8}\)
`b)`
\(\dfrac{17}{15}\div\dfrac{4}{3}\)
`=`\(\dfrac{17}{15}\cdot\dfrac{3}{4}\)
`=`\(\dfrac{17}{3\cdot5}\cdot\dfrac{3}{4}\)
`=`\(\dfrac{17}{5}\cdot\dfrac{1}{4}\)
`=`\(\dfrac{17}{20}\)
`c)`
\(4\dfrac{1}{5}\div\left(-2\dfrac{4}{5}\right)\)
`=`\(4\dfrac{1}{5}\cdot\left(-\dfrac{5}{14}\right)\)
`=`\(\dfrac{21}{5}\cdot\left(-\dfrac{5}{14}\right)\)
`=`\(-\dfrac{21}{14}=-\dfrac{3}{2}\)
a) \(\dfrac{-9}{34}\cdot\dfrac{17}{4}\)
\(=\dfrac{-9\cdot17}{34\cdot4}\)
\(=-\dfrac{153}{136}\)
\(=\dfrac{9}{8}\)
b) \(\dfrac{17}{15}:\dfrac{4}{3}\)
\(=\dfrac{17}{15}\cdot\dfrac{3}{4}\)
\(=\dfrac{17\cdot3}{15\cdot4}\)
\(=\dfrac{51}{60}=\dfrac{17}{20}\)
c) \(4\dfrac{1}{5}:\left(-2\dfrac{4}{5}\right)\)
\(=\dfrac{21}{5}:-\dfrac{14}{5}\)
\(=\dfrac{21}{5}\cdot-\dfrac{5}{14}\)
\(=\dfrac{21\cdot-5}{5\cdot14}\)
\(=-\dfrac{105}{70}=\dfrac{3}{2}\)
Cô làm rồi em nhé:
https://olm.vn/cau-hoi/giup-em-voiii.8161766187032
a.\(\dfrac{17}{15}\div\dfrac{4}{3}=\dfrac{17}{20}\)
b.\(\dfrac{-12}{21}\div\dfrac{34}{43}=\dfrac{-86}{119}\)
c.\(\dfrac{-5}{9}\times\dfrac{3}{11}+\dfrac{13}{18}\times\dfrac{3}{11}\)
=\(\dfrac{3}{11}\times(\dfrac{-5}{9}+\dfrac{13}{18})=\dfrac{3}{11}\times\dfrac{1}{6}=\dfrac{1}{22}\)
d.\(\dfrac{-2}{9}\times\dfrac{5}{11}+\dfrac{-16}{9}\times\dfrac{5}{11}=\dfrac{5}{11}\times(\dfrac{-2}{9}+\dfrac{-16}{9})\)
=\(\dfrac{5}{11}\times(-2)=\dfrac{-10}{11}\)
a)= \(\left(\dfrac{4}{9}-\dfrac{17}{18}\right)+\left(\dfrac{17}{14}-\dfrac{5}{7}\right)+\dfrac{11}{125}\)
= \(\dfrac{-1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{11}{125}\)
= 0 + \(\dfrac{11}{125}\)
= \(\dfrac{11}{125}\)
b) \(=\left(1-1\right)+\left(\dfrac{-1}{2}-\dfrac{1}{2}\right)+\left(2-2\right)\) +
\(\left(\dfrac{-2}{3}-\dfrac{1}{3}\right)+\left(3-3\right)+\left(\dfrac{-3}{4}-\dfrac{1}{4}\right)\) + 4
= 0 + (-1) + 0 + (-1) + 0 + (-1) + 4
= -1
c) = \(\dfrac{1}{3}.\dfrac{14}{25}-\dfrac{1}{2}.\dfrac{14}{25}\)
= \(\dfrac{14}{25}.\left(\dfrac{1}{3}-\dfrac{1}{2}\right)\)
= \(\dfrac{14}{25}.\left(\dfrac{-1}{6}\right)\)
= \(\dfrac{-7}{75}\)
d) = \(\left(\dfrac{3}{7}+\dfrac{4}{7}\right)+\left(\dfrac{5}{13}-\dfrac{18}{13}\right)\)
= 1 + (-1)
= 0
A = \(\dfrac{\dfrac{2022}{1}+\dfrac{2021}{2}+\dfrac{2020}{3}+...+\dfrac{1}{2022}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}}\)
Xét TS = \(\dfrac{2022}{1}\) + \(\dfrac{2021}{2}\) \(\dfrac{2020}{3}\) +... + \(\dfrac{1}{2022}\)
TS = (1 + \(\dfrac{2021}{2}\)) + (1 + \(\dfrac{2020}{3}\)) + ... + ( 1 + \(\dfrac{1}{2022}\)) + 1
TS = \(\dfrac{2023}{2}\) + \(\dfrac{2023}{3}\) +...+ \(\dfrac{2023}{2022}\) + \(\dfrac{2023}{2023}\)
TS = 2023.(\(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) +...+ \(\dfrac{1}{2023}\))
A = \(\dfrac{2023.\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\right)}{\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\right)}\)
A = 2023
\(a)\dfrac{-11}{12}và\dfrac{17}{-18}\) \(\Leftrightarrow\dfrac{-11}{12}và\dfrac{-17}{18}\) \(\Leftrightarrow\dfrac{-33}{36}và\dfrac{-34}{36}\)
Ta thấy rằng : \(-33>-34\Rightarrow\dfrac{-33}{36}>\dfrac{-34}{36}\)
Hay : \(\dfrac{-11}{12}>\dfrac{17}{-18}\)
\(b)\dfrac{-14}{-21}và\dfrac{-60}{-72}\)
Ta có : \(\dfrac{-14}{-21}\text{=}\dfrac{-14:-7}{-21:-7}\text{=}\dfrac{2}{3}\text{=}\dfrac{4}{6}\)
\(\dfrac{-60}{-72}\text{=}\dfrac{-60:-12}{-72:-12}=\dfrac{5}{6}\)
Do đó : \(\dfrac{-14}{-21}< \dfrac{-60}{-72}\)
\(c)\dfrac{2135}{13790}và\dfrac{4}{3}\)
Xét phân số : \(\dfrac{2135}{13790}\) ta thấy rằng : \(tử< mẫu\left(2135< 13790\right)\)
\(\Rightarrow\dfrac{2135}{13790}< 1\)
Xét phân số : \(\dfrac{4}{3}có\) : \(tử>mẫu\left(4>3\right)\)
\(\Rightarrow\dfrac{4}{3}>1\)
Do đó : \(\dfrac{2135}{13790}< \dfrac{4}{3}\)
\(d)\dfrac{2022}{2021}và\dfrac{10}{9}\)
Ta thấy rằng : \(\dfrac{2022}{2021}-\dfrac{1}{2021}\text{=}1\)
\(\dfrac{10}{9}-\dfrac{1}{9}\text{=}1\)
Mà : \(\dfrac{1}{9}>\dfrac{1}{2021}\)
\(\Rightarrow\dfrac{2022}{2021}< \dfrac{10}{9}\)
\(e)\dfrac{35}{36}và\dfrac{16}{17}\)
Ta có : \(\dfrac{35}{36}+\dfrac{1}{36}\text{=}1\)
\(\dfrac{16}{17}+\dfrac{1}{17}\text{=}1\)
Mà : \(\dfrac{1}{36}< \dfrac{1}{17}\)
\(\Rightarrow\dfrac{35}{36}>\dfrac{16}{17}\)
\(f)-1,3< -1,2\)
a) Ta có:
\(-\dfrac{11}{12}=\dfrac{1}{12}-1\)
\(-\dfrac{17}{18}=\dfrac{1}{18}-1\)
Mà: \(\dfrac{1}{12}>\dfrac{1}{18}\)
Hay: \(\dfrac{1}{12}-1>\dfrac{1}{18}-1\Rightarrow-\dfrac{11}{12}>-\dfrac{17}{18}\)
b) Ta có:
\(\dfrac{-14}{-21}=\dfrac{2}{3}=\dfrac{4}{6}\)
\(\dfrac{-60}{-72}=\dfrac{5}{6}\)
Mà: \(5>4\Rightarrow\dfrac{-60}{-72}>\dfrac{-14}{-21}\)
c) Ta có:
\(\dfrac{2135}{13790}=\dfrac{61}{394}< 1\) (tử nhỏ hơn mẫu)
\(\dfrac{4}{3}>1\) (tử lớn hơn mẫu)
Ta có: \(\dfrac{61}{394}< \dfrac{4}{3}\Rightarrow\dfrac{2135}{13790}< \dfrac{4}{3}\)
d) Ta có:
\(\dfrac{2022}{2021}=\dfrac{1}{2021}+1\)
\(\dfrac{10}{9}=\dfrac{1}{9}+1\)
Ta thấy: \(\dfrac{1}{2021}< \dfrac{1}{9}\Rightarrow\dfrac{1}{2021}+1< \dfrac{1}{9}+1\)
Hay \(\dfrac{2022}{2021}< \dfrac{10}{9}\)
e) Ta có:
\(\dfrac{35}{36}=1-\dfrac{1}{36}\)
\(\dfrac{16}{17}=1-\dfrac{1}{17}\)
Ta có: \(\dfrac{1}{36}< \dfrac{1}{17}\Rightarrow1-\dfrac{1}{36}>1-\dfrac{1}{17}\)
Hay \(\dfrac{35}{36}>\dfrac{16}{17}\)
f) Ta có: \(1,3>1,2\)
\(\Rightarrow-1,3< -1,2\)