B=\(\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.....\dfrac{20^2}{19.21}\)
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S
5 tháng 2 2023
\(\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.....\dfrac{99^2}{98.100}\)
\(=\dfrac{2.2.3.3.4.4.....99.99}{1.3.2.4.3.5.....98.100}\)
\(=\dfrac{2.3.4.....99}{1.2.3.4.....98}.\dfrac{2.3.4.....99}{3.4.5.....100}\)
\(=\dfrac{99}{98}\cdot\dfrac{2}{100}\)
\(=\dfrac{99}{4900}\)
ID
8 tháng 5 2022
\(\dfrac{4}{3}\times\dfrac{9}{8}\times\dfrac{16}{15}\times\dfrac{25}{24}=\dfrac{5}{3}\)
8 tháng 5 2022
`(2^2)/(1 . 3) . (3^2)/(2 . 4) . (4^2)/(3 . 5) . (5^2)/(4 . 6)`
`= 4/3 . 9/8 . 16/15 . 25/24 = 5/3`
NQ
1
LG
17 tháng 9 2017
Ta có:B = \(\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}......\dfrac{98.100}{99^2}\)
\(=\dfrac{1.2.3......98}{2.3.4......99}.\dfrac{3.4.5.....100}{2.3.4.....99}=\dfrac{1}{99}.\dfrac{100}{2}=\dfrac{100}{198}\)
Vậy B = \(\dfrac{100}{198}\)
KS
3
H
5 tháng 4 2021
Trước hết ta chứng minh (a-1)(a+1) + 1 = a^2 (*)
Thật vậy VT = (a-1)(a+1)+1=(a-1)a + a-1 +1 = a^2-a+a=a^2 =VP
Áp dụng (*) ta có:
\(A=\dfrac{1\cdot3+2}{2^2}+\dfrac{2\cdot4+2}{3^2}+...+\dfrac{2009\cdot2011+2}{2010^2}\\ =\dfrac{2^2+1}{2^2}+\dfrac{3^2+1}{3^2}+...+\dfrac{2010^2+1}{2010^2}=2009+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2010^2}\\ < 2009+\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2009\cdot2010}\\ =2009+\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{2009}-\dfrac{1}{2010}=2010-\dfrac{1}{2010}< 2020< 2011\)
29 tháng 3 2023
a: \(A=\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{2022\cdot2024}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2022}-\dfrac{1}{2024}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{1011}{2024}=\dfrac{1011}{4848}< \dfrac{1}{4}\)
b: \(B=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2013\cdot2015}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2013}-\dfrac{1}{2015}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2014}{2015}=\dfrac{1007}{2015}< \dfrac{1}{2}\)
ND
0
MV
6 tháng 8 2017
1.
a,
\(\left(\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}+...+\dfrac{2}{19\cdot21}\right)\cdot462-\left[2,04:\left(x+1,05\right)\right]:0,12=19\\ \left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)\cdot462-\left[2,04:\left(x+1,05\right)\right]:0,12=19\\ \left(\dfrac{1}{11}-\dfrac{1}{21}\right)\cdot462-\left[2,04:\left(x+1,05\right)\right]:0,12=19\\ \dfrac{10}{231}\cdot462-\left[2,04:\left(x+1,05\right)\right]:0,12=19\\ 20-\left[2,04:\left(x+1,05\right)\right]:0,12=19\\ \left[2,04:\left(x+1,05\right)\right]:0,12=1\\ 2,04:\left(x+1,05\right)=0,12\\ x+1,05=17\\ x=15,95\)
b,
\(\dfrac{1}{24\cdot25}+\dfrac{1}{25\cdot26}+...+\dfrac{1}{29\cdot30}+x:\dfrac{1}{3}=-4\\ \dfrac{1}{24}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{26}+...+\dfrac{1}{29}-\dfrac{1}{30}+x\cdot3=-4\\ \dfrac{1}{24}-\dfrac{1}{30}+x\cdot3=-4\\ \dfrac{1}{120}+x\cdot3=-4\\ 3x=\dfrac{-481}{120}\\ x=\dfrac{-481}{360}\)
2.
a,
\(\dfrac{15}{28}-\dfrac{186}{1116}-\dfrac{121}{462}+\dfrac{189}{198}\\ =\dfrac{15}{28}-\dfrac{1}{6}-\dfrac{11}{42}+\dfrac{21}{22}\\ =\dfrac{495}{924}-\dfrac{154}{924}-\dfrac{242}{924}+\dfrac{882}{924}\\ =\dfrac{495-154-242+882}{924}\\ =\dfrac{981}{924}\\ =\dfrac{327}{308}\)
b,
\(\left(1+\dfrac{1}{1\cdot3}\right)\cdot\left(1+\dfrac{1}{2\cdot4}\right)\cdot\left(1+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(1+\dfrac{1}{99\cdot101}\right)\\ =\left(\dfrac{1\cdot3}{1\cdot3}+\dfrac{1}{1\cdot3}\right)\cdot\left(\dfrac{2\cdot4}{2\cdot4}+\dfrac{1}{2\cdot4}\right)\cdot\left(\dfrac{3\cdot5}{3\cdot5}+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(\dfrac{99\cdot101}{99\cdot101}+\dfrac{1}{99\cdot101}\right)\\ =\left(\dfrac{2^2-1}{1\cdot3}+\dfrac{1}{1\cdot3}\right)\cdot\left(\dfrac{3^2-1}{2\cdot4}+\dfrac{1}{2\cdot4}\right)\cdot\left(\dfrac{4^2-1}{3\cdot5}+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(\dfrac{100^2-1}{99\cdot101}+\dfrac{1}{99\cdot101}\right)\)\(=\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot...\cdot\dfrac{100^2}{99\cdot101}\\ =\dfrac{2\cdot2}{1\cdot3}\cdot\dfrac{3\cdot3}{2\cdot4}\cdot\dfrac{4\cdot4}{3\cdot5}\cdot...\cdot\dfrac{100\cdot100}{99\cdot101}\\ =\dfrac{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot...\cdot100\cdot100}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot99\cdot101}\\ =\dfrac{\left(2\cdot3\cdot4\cdot...\cdot100\right)\cdot\left(2\cdot3\cdot4\cdot...\cdot100\right)}{\left(1\cdot2\cdot3\cdot...\cdot99\right)\cdot\left(3\cdot4\cdot5\cdot...\cdot101\right)}\\ =\dfrac{100\cdot2}{1\cdot101}\\ =\dfrac{200}{101}\)
NK
1
B=2.2/1.3 . 3.3/2.4 . 4.4/3.5 ......20.20/19.21
=2.3.4.....20/1.2.3.....19 . 2.3.4....20/3.4.5.....21
=20 . 2/21
=40/21