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\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times...\times\left(1-\dfrac{1}{2023}\right)\\ =\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{2022}{2023}\\ =\dfrac{1}{2023}\)
\(\dfrac{11}{2}\): \(\dfrac{1}{4}\) \(\times\) \(\dfrac{5}{3}\)
= \(\dfrac{11}{2}\) \(\times\) \(\dfrac{4}{1}\) \(\times\) \(\dfrac{5}{3}\)
= 22 \(\times\) \(\dfrac{5}{3}\)
= \(\dfrac{110}{3}\)
\(\dfrac{5}{2}-\dfrac{1}{4}+\dfrac{5}{3}\)
= \(\dfrac{30}{12}-\dfrac{3}{12}+\dfrac{20}{12}\)
= \(\dfrac{7}{12}\)
\(\dfrac{14}{5}\times\dfrac{2}{3}\)+ 5
= \(\dfrac{28}{15}\) + 5
= \(\dfrac{28}{15}\) + \(\dfrac{75}{15}\)
= \(\dfrac{103}{15}\)
`@` `\text {Ans}`
`\downarrow`
`(1/2-1/3+1/4-1/5):(1/4-1/5)`
`=`\(\left(\dfrac{1}{6}+\dfrac{1}{20}\right)\div\dfrac{1}{20}\)
`=`\(\dfrac{1}{20}\div\dfrac{1}{20}+\dfrac{1}{6}\div\dfrac{1}{20}\)
`= 1+10/3`
`= 13/3`
A = (\(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\)): (\(\dfrac{1}{4}\) - \(\dfrac{1}{5}\))
A = ( \(\dfrac{30}{60}\) - \(\dfrac{20}{60}\) + \(\dfrac{15}{60}\) - \(\dfrac{12}{60}\)):(\(\dfrac{5}{20}\) - \(\dfrac{4}{20}\))
A = \(\dfrac{13}{60}\): \(\dfrac{1}{20}\)
A = \(\dfrac{13}{60}\times\dfrac{20}{1}\)
A = \(\dfrac{13}{3}\)
Bài giải
a, \(\left(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}\right)\text{ : }\left(\frac{1}{6}+\frac{1}{10}-\frac{1}{15}\right)=\left(\frac{5}{30}+\frac{3}{30}+\frac{2}{30}\right)\text{ : }\left(\frac{5}{30}+\frac{3}{30}-\frac{2}{30}\right)=\frac{1}{3}-\frac{1}{5}=\frac{2}{15}\)
b, \(\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)\text{ : }\left(\frac{1}{4}-\frac{1}{5}\right)=\left(\frac{60}{120}-\frac{40}{120}+\frac{30}{120}-\frac{24}{120}\right)\text{ : }\left(\frac{5}{20}-\frac{4}{20}\right)=\frac{13}{60}\text{ : }\frac{1}{20}=\frac{13}{3}\)
Ta có :
a, \(\left(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}\right)\text{ : }\left(\frac{1}{6}+\frac{1}{10}-\frac{1}{15}\right)=\left(\frac{5}{30}+\frac{3}{30}+\frac{2}{30}\right)\text{ : }\left(\frac{5}{30}+\frac{3}{30}-\frac{2}{30}\right)=\frac{1}{3}-\frac{1}{5}=\frac{2}{15}\)
b,
\(\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)\text{ : }\left(\frac{1}{4}-\frac{1}{5}\right)=\left(\frac{60}{120}-\frac{40}{120}+\frac{30}{120}-\frac{24}{120}\right)\text{ : }\left(\frac{5}{20}-\frac{4}{20}\right)=\frac{13}{60}\text{ : }\frac{1}{20}=\frac{13}{3}\)
\(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{729.3}\)
\(A=1+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)
=> \(3A=3+1+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^6}\)
=> \(3A-A=3+1+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^6}-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)\)
<=> \(2A=3-\frac{1}{3^7}=\frac{3^8-1}{3^7}\)
=> \(A=\frac{3^8-1}{2.3^7}\)
\(A=\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{1}{4950}=\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{9900}=2.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)=2.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=2.\left(\dfrac{1}{2}-\dfrac{1}{100}\right)=2.\dfrac{49}{100}=\dfrac{49}{50}\)
5/2 - 1/4 + 5/3
= 10/4 - 1/4 + 5/3
= 9/4 + 5/3
= 27/12 + 20/12
= 47/12
11/2 : 1/4 x 5/3
= 11/2 x 4/1 x 5/3
= 44/2 x 5/3
= 220/6
= 110/3
14/5 x 2/3 + 5
= 28/15 + 5
= 28/15 + 75/15
= 103/15
\(A=\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{2021}\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{2022}{2021}\)
\(=\frac{2022}{2}\)
\(=1011\)
A= 1* (1/2+1/3+1/4+...+2021)
A= 1/2+1/3+1/4+...+2021
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