Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1: \(S=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{101}{100}=\dfrac{101}{2}\)
2: \(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2006}{2007}=\dfrac{1}{2007}\)
1. \(A=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}=\dfrac{2}{4}=\dfrac{1}{2}\)
2. \(B=\dfrac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\dfrac{1}{5}\)
3.\(C=\dfrac{2^2.3^2.\text{4^2.5^2}.5^2}{1.2^2.3^2.4^2.5.6^2}=\dfrac{125}{36}\)
4.D=\(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right).\dfrac{4}{9}.\dfrac{1}{16}=\dfrac{19}{30}.\dfrac{1}{36}=\dfrac{19}{1080}\)
a) Ta có
S = \(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{n.\left(n+1\right).\left(n+2\right)}\)
2S = \(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{n.\left(n+1\right).\left(n+2\right)}\)
2S = \(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right).\left(n+2\right)}\)2S = \(\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right).\left(n+2\right)}\)
S = \(\dfrac{1}{4}-\dfrac{1}{\left(n+1\right).\left(n+2\right):2}\)
b) A = \(1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{99}\)
A = \(2-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
A = \(2-\dfrac{1}{99}\)
A = \(\dfrac{197}{99}\)
c) Ta có
B = \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\)
B = \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
B = \(1-\dfrac{1}{100}\)
B = \(\dfrac{99}{100}\)
d) Ta có
C = \(\dfrac{99}{1}+\dfrac{98}{2}+\dfrac{97}{3}+...+\dfrac{1}{99}\)
C = \(1+\left(1+\dfrac{98}{2}\right)+\left(1+\dfrac{97}{3}\right)+...+\left(1+\dfrac{1}{99}\right)\)
C = \(1+50+\dfrac{100}{3}+...+\dfrac{100}{99}\)
C = 51 + 100(\(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{99}\))
Đặt D = \(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{99}\)
D = \(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{98}-\dfrac{1}{99}\)
D = \(\dfrac{1}{2}-\dfrac{1}{99}\)
D = \(\dfrac{97}{198}\)
=> C = 51 + 100.\(\dfrac{97}{198}\)
C = 51 + \(\dfrac{4850}{99}\)
C = \(\dfrac{9899}{99}\)
Đây là bài làm của mình sai thì nx nha
Kiểm tra lại đề xem thừa số cuối có đúng quy luật của dãy không.
a)
\(A=\dfrac{3}{4}.\dfrac{8}{9}...\dfrac{9999}{10000}\)
\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}...\dfrac{99.101}{100.100}\)
\(=\dfrac{1.2...99}{2.3...100}.\dfrac{3.4...101}{2.3...100}\)
\(=\dfrac{1}{100}.\dfrac{101}{2}\)
\(=\dfrac{101}{200}\)
a) 15 - 3x = 6
=> 3x = 15 - 6
=> 3x = 9
=> x = 9 : 3 = 3
b) \(\dfrac{2}{3}-\dfrac{4}{3}x=\dfrac{1}{2}\Rightarrow\dfrac{4}{3}x=\dfrac{2}{3}-\dfrac{1}{2}\Rightarrow\dfrac{4}{3}x=\dfrac{1}{6}\Rightarrow x=\dfrac{1}{6}:\dfrac{4}{3}\Rightarrow x=\dfrac{3}{24}=\dfrac{1}{8}\)c) 319.(x - 12 ) = 4 . 320
=> x - 12 = 4 . 3
=> x - 12 = 12
=> x = 12 + 12 = 24
d) 3.(x - 4) = 2^2 . 3^3
=> x - 4 = 2^2 . 3^2
=> x - 4 = 36
=> x = 36 + 4 = 40
a)\(15-3x=6\Rightarrow3x=9\Rightarrow x=3\)
b) \(\dfrac{2}{3}-\dfrac{4}{3}x=\dfrac{1}{2}\Rightarrow\dfrac{4}{3}x=\dfrac{1}{6}\Rightarrow x=\dfrac{1}{8}\)
c) \(3^{19}.\left(x-12\right)=4.3^{20}\Rightarrow x-12=12\Rightarrow x=24\)
d)\(3\left(x-4\right)=2^2.3^3\Rightarrow3x-12=108\Rightarrow3x=120\Rightarrow x=40\)
\(\dfrac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\)
\(=\dfrac{11.3^{29}-\left(3^2\right)^{15}}{2^2.3^{28}}\)
\(=\dfrac{11.3^{29}-3^{30}}{2^2.3^{28}}\)
\(=\dfrac{3^{29}\left(11-3\right)}{2^2.3^{28}}\)
\(=\dfrac{3^{29}.2^3}{2^2.3^{28}}\)
\(=\dfrac{3.2}{1.1}=6\)
Biến đổi thừa số tổng quát: \(1+\dfrac{1}{\left(k-1\right)\left(k+1\right)}\) \(=\dfrac{\left(k-1\right)\left(k+1\right)+1}{\left(k-1\right)\left(k+1\right)}\) \(=\dfrac{k^2}{\left(k-1\right)\left(k+1\right)}\).
Do đó \(1+\dfrac{1}{1.3}=\dfrac{2^2}{1.3}\), \(1+\dfrac{1}{2.4}=\dfrac{3^2}{2.4}\), \(1+\dfrac{1}{3.5}=\dfrac{4^2}{3.5}\),..., \(1+\dfrac{1}{2018.2020}=\dfrac{2019^2}{2018.2020}\), \(1+\dfrac{1}{2019.2021}=\dfrac{2020^2}{2019.2021}\). Từ đó suy ra \(\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)...\left(1+\dfrac{1}{2019.2021}\right)\)
\(=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.\dfrac{5^2}{4.6}.\dfrac{6^2}{5.7}...\dfrac{2019^2}{2018.2020}.\dfrac{2020^2}{2019.2021}\)
\(=\dfrac{2.2020}{2021}=\dfrac{4040}{2021}\)
Bài 2:
a: \(=44\cdot82-400+18\cdot44\)
\(=44\cdot100-400=4400-400=4000\)
b: \(=6^2:\left\{780:\left[390-125\cdot49+65\right]\right\}\)
\(=36:\left\{780:\left[-5670\right]\right\}\)
\(=36:\dfrac{-26}{189}=\dfrac{-3402}{13}\)
\(\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot...\cdot\dfrac{59^2}{58\cdot60}\)
\(=\dfrac{2\cdot3\cdot...\cdot59}{1\cdot2\cdot...\cdot58}\cdot\dfrac{2\cdot3\cdot...\cdot59}{3\cdot4\cdot...\cdot60}\)
\(=\dfrac{59}{1}\cdot\dfrac{2}{60}=\dfrac{59}{30}\)
=2.2/1.3 + 3.3/2.4 + 4.4/3.5 + ... + 59.59/58.60
=2.3.4. ... . 59 /1.2.3. ... . 58 + 2.3.4. ... .59 / 3.4.5. ... .60
=(RÚT GỌN ĐI) = 59/1 + 2/60
= 59+1/30
= 59 và 1/30