\(D=\dfrac{1}{1-\dfrac{1}{1-2^2}}+\dfrac{1}{1+\dfrac{1}{1+2^2}}\)
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23 tháng 4 2021 lúc 13:31
a/ \(A=\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{99.100}\)
b/ \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{20}}< 1\)
c/ \(A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{97.99}\)
d/ \(A=\dfrac{2015}{2016}+\dfrac{2016}{2017}+\dfrac{2017}{2018}+\dfrac{2018}{2015}>4\)
a) 0,25-dfrac{2}{3}+1dfrac{1}{4}b) dfrac{3^2}{2}:dfrac{1}{4}+dfrac{3}{4}.2010c) {[(dfrac{1}{25}-0,6)2:dfrac{49}{125}].dfrac{5}{6}}-[(dfrac{-1}{3})+dfrac{1}{2}]d) (-dfrac{1}{2}-dfrac{1^{ }}{3})2:[(dfrac{-5}{36})-(dfrac{-5}{36})0]Mn giúp mk nhé mk gấp quá tí đi học ai làm được mk thả tim và like nhé
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a) 0,25-\(\dfrac{2}{3}\)+1\(\dfrac{1}{4}\)
b) \(\dfrac{3^2}{2}\):\(\dfrac{1}{4}\)+\(\dfrac{3}{4}\).2010
c) {[(\(\dfrac{1}{25}\)-0,6)2:\(\dfrac{49}{125}\)].\(\dfrac{5}{6}\)}-[(\(\dfrac{-1}{3}\))+\(\dfrac{1}{2}\)]
d) (-\(\dfrac{1}{2}\)-\(\dfrac{1^{ }}{3}\))2:[(\(\dfrac{-5}{36}\))-(\(\dfrac{-5}{36}\))0]
Mn giúp mk nhé mk gấp quá tí đi học ai làm được mk thả tim và like nhé
a) \(0,25-\dfrac{2}{3}+1\dfrac{1}{4}\)
\(=\dfrac{1}{4}-\dfrac{2}{3}+\dfrac{5}{4}\)
\(=\dfrac{3}{12}-\dfrac{8}{12}+\dfrac{15}{12}\)
\(=\dfrac{10}{12}\)
\(=\dfrac{5}{6}\)
\(---\)
b) \(\dfrac{3^2}{2}:\dfrac{1}{4}+\dfrac{3}{4}\cdot2010\)
\(=\dfrac{9}{2}\cdot4+\dfrac{3015}{2}\)
\(=18+\dfrac{3015}{2}\)
\(=\dfrac{36}{2}+\dfrac{3015}{2}\)
\(=\dfrac{3051}{2}\)
\(---\)
c) \(\left\{\left[\left(\dfrac{1}{25}-0,6\right)^2:\dfrac{49}{125}\right]\cdot\dfrac{5}{6}\right\}-\left[\left(\dfrac{-1}{3}\right)+\dfrac{1}{2}\right]\)
\(=\left\{\left[\left(-\dfrac{14}{25}\right)^2:\dfrac{49}{125}\right]\cdot\dfrac{5}{6}\right\}-\left[\left(\dfrac{-2}{6}\right)+\dfrac{3}{6}\right]\)
\(=\left\{\left[\dfrac{196}{625}\cdot\dfrac{125}{49}\right]\cdot\dfrac{5}{6}\right\}-\dfrac{1}{6}\)
\(=\left\{\dfrac{4}{5}\cdot\dfrac{5}{6}\right\}-\dfrac{1}{6}\)
\(=\dfrac{4}{6}-\dfrac{1}{6}\)
\(=\dfrac{3}{6}\)
\(=\dfrac{1}{2}\)
\(---\)
d) \(\left(-\dfrac{1}{2}-\dfrac{1}{3}\right)^2:\left[\left(\dfrac{-5}{36}\right)-\left(\dfrac{-5}{36}\right)^0\right]\)
\(=\left(-\dfrac{3}{6}-\dfrac{2}{6}\right)^2:\left[-\dfrac{5}{36}-1\right]\)
\(=\left(-\dfrac{5}{6}\right)^2:\left[-\dfrac{5}{36}-\dfrac{36}{36}\right]\)
\(=\dfrac{25}{36}:\left(\dfrac{-41}{36}\right)\)
\(=\dfrac{25}{36}\cdot\left(\dfrac{-36}{41}\right)\)
\(=-\dfrac{25}{41}\)
#\(Toru\)
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Hãy tính \(\dfrac{C}{D}\). Biết C= \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{48}+\dfrac{1}{49}+\dfrac{1}{50}\) và D= \(\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)
=> D + 49 = (1/49 + 1) + (2/48 + 1) +... (49/1 + 1)
= 50/1 + 50/2 + ... + 50/49
= 50(1/2+1/3+...+1/49) + 50
=> D = 50(1/2 + 1/3 +... + 1/49) + 1
= 50(1/2 + 1/3 +... + 1/49 + 1/50)
=> C/D = 1/50
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24 tháng 2 2022 lúc 11:03
bài 1 : tính
\(B=\dfrac{1}{10.9}+\dfrac{1}{18.13}+\dfrac{1}{26.27}+...+\dfrac{1}{802.405}\)
\(D=\dfrac{1}{2}-\dfrac{1}{2^4}+\dfrac{1}{2^7}-\dfrac{1}{2^{10}}+...-\dfrac{1}{2^{58}}\)
mn giúp mk gấp nhé !!
Bài 1:
a, Cho A dfrac{1}{2^2}+dfrac{1}{4^2}+dfrac{1}{6^2}+....+dfrac{1}{100^2}
Chứng tỏ: A dfrac{1}{2}
b, Cho B dfrac{1}{2}+dfrac{1}{2^2}+dfrac{1}{2^3}+....+dfrac{1}{2^{20}}
Chứng tỏ B 1
c, Cho C dfrac{1}{5}+dfrac{1}{13}+dfrac{1}{14}+dfrac{1}{15}+dfrac{1}{61}+dfrac{1}{62}+dfrac{1}{63}
Chứng tỏ C dfrac{1}{2}
d, Cho D dfrac{1}{2^2}+dfrac{1}{3^2}+dfrac{1}{4^2}+...+dfrac{1}{20^2}
Chứng tỏ D 1
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Bài 1:
a, Cho A = \(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+....+\dfrac{1}{100^2}\)
Chứng tỏ: A <\(\dfrac{1}{2}\)
b, Cho B = \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{20}}\)
Chứng tỏ B < 1
c, Cho C = \(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\)
Chứng tỏ C < \(\dfrac{1}{2}\)
d, Cho D = \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{20^2}\)
Chứng tỏ D < 1
Giải
Ta có : \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{20^2}< \dfrac{1}{19.20}\)
\(\Rightarrow\)D < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{19.20}\)
Nhận xét: \(\dfrac{1}{1.2}=1-\dfrac{1}{2};\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3};\dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4};...;\dfrac{1}{19.20}=\dfrac{1}{19}-\dfrac{1}{20}\)
\(\Rightarrow\) D< 1- \(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
D< 1 - \(\dfrac{1}{20}\)
D< \(\dfrac{19}{20}\)<1
\(\Rightarrow\)D< 1
Vậy D=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{5^2}\)<1
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A=\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)
A=\(\dfrac{1}{2^2.1}+\dfrac{1}{2^2.2^2}+\dfrac{1}{3^2.2^2}+...+\dfrac{1}{50^2.2^2}\)
A=\(\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)\)
\(A=\dfrac{1}{2^2}\left(1+\dfrac{1}{2.2}+\dfrac{1}{3.3}+...+\dfrac{1}{50.50}\right)\)
Ta có :
\(\dfrac{1}{2.2}< \dfrac{1}{1.2};\dfrac{1}{3.3}< \dfrac{1}{2.3};\dfrac{1}{4.4}< \dfrac{1}{3.4};...;\dfrac{1}{50.50}< \dfrac{1}{49.50}\)
\(\Rightarrow A< \dfrac{1}{2^2}\left(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\right)\)Nhận xét :
\(\dfrac{1}{1.2}< 1-\dfrac{1}{2};\dfrac{1}{2.3}< \dfrac{1}{2}-\dfrac{1}{3};...;\dfrac{1}{49.50}< \dfrac{1}{49}-\dfrac{1}{50}\)
\(\Rightarrow A< \dfrac{1}{2^2}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
A<\(\dfrac{1}{2^2}\left(1-\dfrac{1}{50}\right)\)
A<\(\dfrac{1}{4}.\dfrac{49}{50}\)<1
A<\(\dfrac{49}{200}< \dfrac{1}{2}\)
\(\Rightarrow A< \dfrac{1}{2}\)
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!!!
1. Tìm lim una. u_ndfrac{1}{2^2-1}+dfrac{1}{3^2-1}+...+dfrac{1}{n^2-1}b. u_ndfrac{1}{1.2}+dfrac{1}{2.3}+dfrac{1}{3.4}+...+dfrac{1}{nleft(n+1right)}c.u_ndfrac{1}{1}+dfrac{1}{1+2}+...+dfrac{1}{1+2+...+n}d. u_nleft[left(1-dfrac{1}{2^2}right)left(1-dfrac{1}{3^2}right)...left(1-dfrac{1}{n^2}right)right]Ai giúp mk với hoặc gợi ý cho mik cx đc . Tks nhiều
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1. Tìm lim un
a. \(u_n=\dfrac{1}{2^2-1}+\dfrac{1}{3^2-1}+...+\dfrac{1}{n^2-1}\)
b. \(u_n=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n+1\right)}\)
c.\(u_n=\dfrac{1}{1}+\dfrac{1}{1+2}+...+\dfrac{1}{1+2+...+n}\)
d. \(u_n=\left[\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{n^2}\right)\right]\)
Ai giúp mk với hoặc gợi ý cho mik cx đc . Tks nhiều
a.
\(u_n=\dfrac{1}{\left(2-1\right)\left(2+1\right)}+\dfrac{1}{\left(3-1\right)\left(3+1\right)}+...+\dfrac{1}{\left(n-1\right)\left(n+1\right)}\)
\(=\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(n-2\right)n}+\dfrac{1}{\left(n-1\right)\left(n+1\right)}\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{n-2}-\dfrac{1}{n}+\dfrac{1}{n-1}-\dfrac{1}{n+1}\right)\)
\(=\dfrac{1}{2}\left(1+\dfrac{1}{2}-\dfrac{1}{n}-\dfrac{1}{n+1}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{3}{2}-\dfrac{1}{n}-\dfrac{1}{n+1}\right)\)
\(\Rightarrow\lim u_n=\lim\left(\dfrac{1}{2}\left(\dfrac{3}{2}-\dfrac{1}{n}-\dfrac{1}{n+1}\right)\right)=\dfrac{1}{2}.\dfrac{3}{2}=\dfrac{3}{4}\)
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b.
\(u_n=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n+1\right)}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\)
\(=1-\dfrac{1}{n+1}\)
\(\Rightarrow\lim u_n=\lim\left(1-\dfrac{1}{n+1}\right)=1\)
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c.
\(1+2+...+n=\dfrac{n\left(n+1\right)}{2}\)
\(\Rightarrow\dfrac{1}{1+2+...+n}=\dfrac{2}{n\left(n+1\right)}=\dfrac{2}{n}-\dfrac{2}{n+1}\)
\(\Rightarrow u_n=1+\dfrac{2}{2}-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{4}+...+\dfrac{2}{n}-\dfrac{2}{n+1}\)
\(=1+1-\dfrac{2}{n+1}=2-\dfrac{2}{n+1}\)
\(\Rightarrow\lim u_n=\lim\left(2-\dfrac{2}{n+1}\right)=2\)
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Xem thêm câu trả lời
Tính :
1) C = \(\left(\dfrac{1}{200^2}-1\right)\left(\dfrac{1}{199^2}-1\right)...\left(\dfrac{1}{101^2}-1\right)\)
2) \(D=\dfrac{1}{1-\dfrac{1}{1-2^{-1}}}+\dfrac{1}{1+\dfrac{1}{1+2^{-1}}}\)
\(C=\left(\dfrac{1}{200^2}-1\right)\left(\dfrac{1}{199^2-1}\right)...\left(\dfrac{1}{101^2-1}\right)\)
\(C=\dfrac{1-200^2}{200^2}.\dfrac{1-199^2}{199^2}.\dfrac{1-198^2}{198^2}...\dfrac{1-101^2}{101^2}\)
\(C=\dfrac{\left(1-200\right)\left(1+200\right)}{200^2}.\dfrac{\left(1-199\right)\left(1+199\right)}{199^2}...\dfrac{\left(1-100\right)\left(1+100\right)}{100^2}.\dfrac{\left(1-101\right)\left(1+101\right)}{101^2}\) \(C=\dfrac{-199.201}{200.200}.\dfrac{-198.200}{199.199}.\dfrac{-197.199}{198.198}...\dfrac{-99.101}{100.100}.\dfrac{-100.102}{101.101}\)
\(C=\dfrac{199.201}{200.200}.\dfrac{198.200}{199.199}.\dfrac{197.199}{198.198}...\dfrac{99.101}{100.100}.\dfrac{100.102}{101.101}\)
\(\Rightarrow C=\dfrac{200}{2.101}=\dfrac{201}{202}\)
Câu 2 mik chịu r sorry:(
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1. So sánh
a) \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\) và B= \(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{13}{60}\)
b) \(C=\dfrac{2019}{2021}+\dfrac{2021}{2022}\) và \(D=\dfrac{2020+2022}{2019+2021}.\dfrac{3}{2}\)
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.
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D=\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2005}\right)\)
E=\(\dfrac{1^2}{1.3}.\dfrac{2^2}{2.4}.\dfrac{3^2}{3.5}....\dfrac{999^2}{999.1000}.\dfrac{1000^2}{1000.1001}\)
Ta có: D\(=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2005}\right)\)
\(\Leftrightarrow D=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2004}{2005}=\dfrac{1.2.3...2004}{2.3.4...2005}=\dfrac{1}{2005}\)
Ta có: \(E=\dfrac{1^2}{1.3}.\dfrac{2^2}{2.4}.\dfrac{3^2}{3.5}...\dfrac{999^2}{999.1000}.\dfrac{1000^2}{1000.1001}=\dfrac{\left(1.2.3.4...1000\right)\left(1.2.3.4...1000\right)}{\left(1.2.3....1000\right)\left(3.4.5....1001\right)}=\dfrac{2}{1001}\)
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Bài 1.(2,5 điểm)Tìm x, biết:a) left(2dfrac{1}{3}+3dfrac{1}{2}right).x-4dfrac{1}{6}+3dfrac{1}{2}b) left(1dfrac{1}{3}+3dfrac{1}{2}right).x4dfrac{1}{6}-3dfrac{1}{2}c) dfrac{1}{3}-dfrac{7}{8}.xdfrac{1}{4}d) dfrac{3}{2}.x+dfrac{1}{7}dfrac{7}{8}.dfrac{64}{49}e) 5dfrac{1}{2}-left(dfrac{1}{4}.x+dfrac{2}{5}right)25%
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Bài 1.(2,5 điểm)Tìm x, biết:
a) \(\left(2\dfrac{1}{3}+3\dfrac{1}{2}\right).x=-4\dfrac{1}{6}+3\dfrac{1}{2}\)
b) \(\left(1\dfrac{1}{3}+3\dfrac{1}{2}\right).x=4\dfrac{1}{6}-3\dfrac{1}{2}\)
c) \(\dfrac{1}{3}-\dfrac{7}{8}.x=\dfrac{1}{4}\)
d) \(\dfrac{3}{2}.x+\dfrac{1}{7}=\dfrac{7}{8}.\dfrac{64}{49}\)
e) \(5\dfrac{1}{2}-\left(\dfrac{1}{4}.x+\dfrac{2}{5}\right)=25\%\)
c: Ta có: \(\dfrac{1}{3}-\dfrac{7}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow x\cdot\dfrac{7}{8}=\dfrac{1}{12}\)
\(\Leftrightarrow x=\dfrac{1}{12}\cdot\dfrac{8}{7}=\dfrac{2}{21}\)
d: Ta có: \(\dfrac{3}{2}x+\dfrac{1}{7}=\dfrac{7}{8}\cdot\dfrac{64}{49}\)
\(\Leftrightarrow x\cdot\dfrac{3}{2}=1\)
hay \(x=\dfrac{2}{3}\)
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