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2 tháng 6 2018
a) \(A=98+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\)(có 98 phân số nên ta cộng 1 vào mỗi phân số)
\(A=\left(\frac{1}{2}+1\right)+\left(\frac{1}{3}+1\right)+...+\left(\frac{1}{99}+1\right)\)
\(A=\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}\)
Và \(B=\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}}{\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}}=1\)
b) \(A=2018+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\)(có 2018 phân số nên ta cộng 1 vào mỗi phân số)
\(A=\left(\frac{1}{2}+1\right)+\left(\frac{1}{3}+1\right)+...+\left(\frac{1}{2019}+1\right)\)
\(A=\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}\)
Và \(B=\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}}{\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}}=1\)
c) \(A=\frac{99}{1}+\frac{98}{2}+...+\frac{1}{99}\)
\(A=99+\frac{98}{2}+...+\frac{1}{99}\)(có 98 phân số nên ta cộng 1 vào từng phân số)
\(A=\left(\frac{98}{2}+1\right)+\left(\frac{97}{3}+1\right)+...+\left(\frac{1}{99}+1\right)+1\)
\(A=\frac{100}{2}+\frac{100}{3}+...+\frac{100}{99}+1\)
\(A=100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)\)
Và \(B=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\)
\(\Rightarrow\frac{A}{B}=\frac{100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}}=100\)
DL
2 tháng 6 2018
a)\(B=\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+...+\frac{100}{99}\)
\(B=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{4}\right)+...+\left(1+\frac{1}{99}\right)\)
\(\Rightarrow B=\left(1+1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}\right)\)
\(\Rightarrow B=98+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}\)
\(\Rightarrow A:B=\frac{A}{B}=\frac{98+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}}{98+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}}=1.\)
Vậy \(A:B=1.\)
b)\(B=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{4}\right)+...+\left(1+\frac{1}{2019}\right)\)
\(\Rightarrow B=\left(1+1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right)\)
\(\Rightarrow B=2018+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\)
\(\Rightarrow A:B=\frac{A}{B}=\frac{2018+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}}{2018+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}}=1.\)
Vậy \(A:B=1.\)
c)\(A=\left(1+1+...+1\right)+\frac{98}{2}+\frac{97}{3}+...+\frac{2}{98}+\frac{1}{99}\)
\(A=\left(1+\frac{98}{2}\right)+\left(1+\frac{97}{3}\right)+...+\left(1+\frac{2}{98}\right)+\left(1+\frac{1}{99}\right)\)
\(A=\frac{100}{2}+\frac{100}{3}+...+\frac{100}{98}+\frac{100}{99}\)
\(A=100\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{98}+\frac{1}{99}\right)\)
\(\Rightarrow A:B=\frac{A}{B}=\frac{100\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{98}+\frac{1}{99}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{98}+\frac{1}{99}}=1.\)
Vậy \(A:B=1.\)
10 tháng 9 2020
số lượng số hạng của dãy số là
( 2021 - 2 ) : 1 + 1 = 2020
tổng của dãy số là
( 2021 + 2) x 2020 : 2 = 2043230
vậy A = \(\frac{1}{2043230}\)
KK
2
HH
5 tháng 4 2020
\(\frac{1}{1}+2+\frac{1}{1}+2+3+\frac{1}{1} +2+3+4+...+\frac{1}{1}+2+3......2019\)
Ta có : \(\frac{2}{2}+\left(\frac{1}{2}\right)+\frac{2}{2}+\left(1+2+3\right)+....+\frac{2}{2}+\left(1+2+....+50\right)\)
\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.....+\frac{2}{2550}\)
\(=\frac{2}{2}.3+\frac{2}{3}.4+....+\frac{2}{50}.51\)
\(=2.\left(\frac{1}{2}.3+\frac{1}{3}.4+.....+\frac{1}{50}.51\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{50}-\frac{1}{51}\right)\)
\(=2.\left(1-\frac{1}{51}\right)\)
\(=2.\frac{50}{51}\)
\(=\frac{100}{51}\)
Hmmm , kh bt có đúng kh nhỉ ???
Nếu kh đúng chỗ nào mong m.n chỉ ạ
:>>
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
3 tháng 8 2023
Ok em, để olm.vn giúp em nhá:
A = \(\dfrac{1}{2}\):3 + \(\dfrac{1}{3}\):4 + \(\dfrac{1}{4}\):5+...+\(\dfrac{1}{2018}\):2019 + \(\dfrac{1}{2019}\): 2020
A=\(\dfrac{1}{2}\times\dfrac{1}{3}+\dfrac{1}{3}\times\dfrac{1}{4}+\dfrac{1}{4}\times\dfrac{1}{5}+..+\dfrac{1}{2018}\times\dfrac{1}{2019}+\dfrac{1}{2019}\times\dfrac{1}{2020}\)
A = \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\)+....+ \(\dfrac{1}{2018}\) - \(\dfrac{1}{2019}\)+ \(\dfrac{1}{2019}\) - \(\dfrac{1}{2020}\)
A = \(\dfrac{1}{2}\) - \(\dfrac{1}{2020}\)
A = \(\dfrac{1009}{2020}\)
NA
6 tháng 8 2018
A=(2/2-1/2) . (3/3-1/3) . ( 4/4 - 1/4 ) . (5/5 - 1/5) .... (2018/2018-1/2018). (2019/2019 - 1/2019)
A= 1/2 . 2/3 . 3/4 . 5/5 ..... 2017/2018 . 2018/2019
A= 1/2019
CT
6 tháng 8 2018
A = 1-1/2 . 1- 1/3 . 1-1/4 . 1-1/5 . ... . 1-1/2018 . 1-1/2019
= 0 . 0 . 0 . 0 . ... . 0 . 0.
= 0
XO
12 tháng 8 2019
B = \(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}...+\frac{1}{1+2+3+...+2019}\)
= \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{2019\times1010}\)
= \(2\times\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{2019\times2020}\right)\)
= \(2\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{2019\times2020}\right)\)
= \(2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2019}-\frac{1}{2020}\right)\)
= \(2\times\left(\frac{1}{2}-\frac{1}{2020}\right)\)
\(=2\times\frac{1009}{2020}\)
\(=\frac{1009}{1010}< \frac{1010}{1010}=1\)
\(\Rightarrow B< 1\)
DT
0
TD
4
30 tháng 6 2018
\(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times...\times\left(1-\frac{1}{2018}\right)\times\left(1-\frac{1}{2019}\right)\)
\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times...\times\frac{2017}{2018}\times\frac{2018}{2019}\)
\(=\frac{1\times2\times3\times...\times2017\times2018}{2\times3\times4\times...\times2018\times2019}\)
\(=\frac{1}{2019}\)
BK
1