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11 tháng 5 2023
\(A>\dfrac{2^{2018}}{2^{2018}+3^{2019}+5^{2020}}+\dfrac{3^{2019}}{2^{2018}+3^{2019}+5^{2020}}+\dfrac{5^{2020}}{5^{2020}+2^{2018}+3^{2019}}=1\)
\(B< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2019\cdot2020}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2019}-\dfrac{1}{2020}\)
=>B<1
=>A>B
26 tháng 3 2020
M =1/1.2+1/2.3+1/3.4+.......+1/2019.2020
=1-1/2+1/2-1/3+1/3-1/4+.......-1/2019+1/2019-1/2020
=1-1/2020
=2019/2020
AH
Akai Haruma
Giáo viên
14 tháng 5 2019
Lời giải:
\(B=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{2019.2020}\)
\(\Rightarrow 2B=\frac{2}{1.2}+\frac{2}{3.4}+\frac{2}{5.6}+....+\frac{2}{2019.2020}\)
\(< 1+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{2018.2019}+\frac{1}{2019.2020}\)
\(2B< 1+\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+....+\frac{2019-2018}{2018.2019}+\frac{2020-2019}{2019.2020}\)
\(2B< 1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
\( 2B< 1+\frac{1}{2}-\frac{1}{2020}< 1+\frac{1}{2}\)
\(B< \frac{3}{4}\)
---------------------
Đặt \(2^{2018}=a; 3^{2019}=b; 5^{2020}=c(a,b,c>0)\)
\(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}> \frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1\)
\(\Rightarrow A>1> \frac{3}{4}> B\)
1 tháng 8 2021
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2019}-\dfrac{1}{2020}\\ =\dfrac{1}{1}-\dfrac{1}{2020}=\dfrac{2019}{2020}\)
27 tháng 3 2016
$2015=5.13.31$2015=5.13.31
Ta có: $1.2.....1007=1.2...5....13.....31...1007\text{ chia hết cho }5.13.31=2015$1.2.....1007=1.2...5....13.....31...1007 chia hết cho 5.13.31=2015
$1008.1009.....2004=1008....\left(1010\right)....\left(1014\right)...\left(1023\right)....2004$1008.1009.....2004=1008....(1010)....(1014)...(1023)....2004
$=1008....\left(5.202\right)....\left(13.78\right)....\left(31.33\right)...2004\text{ chia hết cho }5.13.33=2015$=1008....(5.202)....(13.78)....(31.33)...2004 chia hết cho 5.13.33=2015
Do đó tổng 2 số trên chia hết cho 2015.
9 tháng 4 2020
a) Ta có:
1/( 2.3 ) = ( 3 - 2 )/( 2.3 )
= 3/( 2.3 ) - 2/( 2.3 )
= 1/2 - 1/3.
1/( 3.4 ) = ( 4 - 3 )/( 3.4 )
= 4/( 3.4 ) - 3/( 3.4 )
= 1/3 - 1/4.
b)
Ta có:
A = 1/( 5.6 ) + 1/( 6.7 ) + 1/( 7.8 ) + ..... + 1/( 2019.2020 )
A = 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + ..... + 1/2019 - 1/2020
A = 1/5 - 1/2020
A = 403/2020
Vậy A = 403/2020.
NN
9 tháng 4 2020
a) Ta có: \(\frac{1}{2.3}=\frac{3-2}{2.3}=\frac{3}{2.3}-\frac{2}{2.3}=\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{3.4}=\frac{4-3}{3.4}=\frac{4}{3.4}-\frac{3}{3.4}=\frac{1}{3}-\frac{1}{4}\)
b) Ta có: \(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+.......+\frac{1}{2019.2020}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+........+\frac{1}{2019}-\frac{1}{2020}\)
\(=\frac{1}{5}-\frac{1}{2020}=\frac{403}{2020}\)
S
24 tháng 8 2019
\(\frac{1}{1.2}+\frac{1}{2.3}+......+\frac{1}{2019.2020}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..-\frac{1}{2020}=1-\frac{1}{2020}=\frac{2019}{2020}\)
\(\Rightarrow a=\frac{2020}{2019}\)
14 tháng 6 2020
\(A=\frac{4}{1.2}+\frac{4}{2.3}+\frac{4}{3.4}+...+\frac{4}{2019.2020}\)
\(\frac{1}{4}A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)
\(\frac{1}{4}A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(\frac{1}{4}A=1-\frac{1}{2020}=\frac{2019}{2020}\)
\(\Rightarrow A=\frac{2019}{2020}:\frac{1}{4}=\frac{2019}{505}\)
Vậy \(A=\frac{2019}{505}.\)
\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
\(\Rightarrow2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\)
\(2B=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\)
\(2B=\frac{1}{1.2}-\frac{1}{99.100}=\frac{4949}{9900}\)
\(\Rightarrow B=\frac{4949}{9900}:2=\frac{4949}{19800}\)
Vậy \(B=\frac{4949}{19800}.\)
14 tháng 6 2020
\(A=\frac{4}{1\cdot2}+\frac{4}{2\cdot3}+\frac{4}{3\cdot4}+...+\frac{4}{2019\cdot2020}\)
\(A=4\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2018\cdot2019}\right)\)
\(A=4\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(A=4\left(1-\frac{1}{2019}\right)=4\cdot\frac{2018}{2019}\)
Đến đây tự tính
\(B=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{98\cdot99\cdot100}\)
\(B=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{98\cdot99\cdot100}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{99\cdot100}\right)=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)\)
Số hơi bị dữ nên tính nốt nhé