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17 tháng 8 2018
a) \(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{99.101}\)
\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\right)+\left(\frac{1}{2.4}+...+\frac{1}{98.100}\right)\)
\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)+2.\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(=2.\left(1-\frac{1}{101}\right)+2.\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=2\cdot\frac{100}{101}+2\cdot\frac{49}{100}=\frac{200}{101}+\frac{49}{50}\)
câu b mk ko bk! xl bn nha!
18 tháng 8 2018
mk nhầm
...
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{100}\right)\) 1/100)
= 1/2.(1-1/101) + 1/2.(1/2-1/100)
=1/2.100/101 + 1/2.49/100
= 50/101 + 49/200
DX
1
15 tháng 9 2019
M = 1/3.5 + 1/5.7 + 1/7.9 + ... + 1/2017.2019
M = 1/2.(1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/2017 - 1/2019)
M = 1/2.(1/3 - 1/2019)
M = 1/2.224/673
M = 112/673
DL
11 tháng 6 2018
Bài 1:
Ta có:
\(N=\frac{2017+2018}{2018+2019}=\frac{2017}{2018+2019}+\frac{2018}{2018+2019}\)
Do \(\hept{\begin{cases}\frac{2017}{2018+2019}< \frac{2017}{2018}\\\frac{2018}{2018+2019}< \frac{2018}{2019}\end{cases}\Rightarrow\frac{2017}{2018+2019}+\frac{2018}{2018+2019}< \frac{2017}{2018}+\frac{2018}{2019}}\)
\(\Leftrightarrow N< M\)
Vậy \(M>N.\)
Bài 2:
Ta có:
\(A=\frac{2017}{987653421}+\frac{2018}{24681357}=\frac{2017}{987654321}+\frac{2017}{24681357}+\frac{1}{24681357}\)
\(B=\frac{2018}{987654321}+\frac{2017}{24681357}=\frac{1}{987654321}+\frac{2017}{987654321}+\frac{2017}{24681357}\)
Do \(\hept{\begin{cases}\frac{2017}{987654321}+\frac{2017}{24681357}=\frac{2017}{987654321}+\frac{2017}{24681357}\\\frac{1}{24681357}>\frac{1}{987654321}\end{cases}}\)
\(\Rightarrow\frac{2017}{987654321}+\frac{2017}{24681357}+\frac{1}{24681357}>\frac{1}{987654321}+\frac{2017}{987654321}+\frac{2017}{24681357}\)
\(\Leftrightarrow A>B\)
Vậy \(A>B.\)
Bài 3:
\(\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}+\frac{2019}{2016}=1-\frac{1}{2017}+1-\frac{1}{2018}+1-\frac{1}{2019}+1+\frac{3}{2016}\)
\(=1+1+1+1-\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}+\frac{3}{2016}\)
\(=4-\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}-\frac{3}{2016}\right)\)
Do \(\hept{\begin{cases}\frac{1}{2017}< \frac{1}{2016}\\\frac{1}{2018}< \frac{1}{2016}\\\frac{1}{2019}< \frac{1}{2016}\end{cases}\Rightarrow\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}< \frac{1}{2016}+\frac{1}{2016}+\frac{1}{2016}=\frac{3}{2016}}\)
\(\Rightarrow\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}-\frac{3}{2016}\)âm
\(\Rightarrow4-\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}-\frac{3}{2016}\right)>4\)
Vậy \(\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}+\frac{2019}{2016}>4.\)
Bài 4:
\(\frac{1991.1999}{1995.1995}=\frac{1991.\left(1995+4\right)}{\left(1991+4\right).1995}=\frac{1991.1995+1991.4}{1991.1995+4.1995}\)
Do \(\hept{\begin{cases}1991.1995=1991.1995\\1991.4< 1995.4\end{cases}}\Rightarrow1991.1995+1991.4< 1991.1995+1995.4\)
\(\Rightarrow\frac{1991.1995+1991.4}{1991.1995+4.1995}< \frac{1991.1995+1995.4}{1991.1995+4.1995}=1\)
\(\Rightarrow\frac{1991.1999}{1995.1995}< 1\)
Vậy \(\frac{1991.1999}{1995.1995}< 1.\)
22 tháng 5 2019
\((2,5\cdot x+2017)\cdot2018=(7,5+2017)\cdot2018\)
\(\Rightarrow(2,5\cdot x+2017)\cdot2018=4085441\)
\(\Rightarrow2,5\cdot x+2017=2024,5\)
\(\Rightarrow2,5x=7,5\)
\(\Rightarrow x=7,5:2,5=3\)
22 tháng 5 2019
\(3\frac{1}{5}+\frac{2}{5}\left[x+\frac{1}{3}\right]=\frac{21}{5}\)
\(\Rightarrow\frac{16}{5}+\frac{2}{5}\left[x+\frac{1}{3}\right]=\frac{21}{5}\)
\(\Rightarrow\frac{2}{5}\left[x+\frac{1}{3}\right]=\frac{21}{5}-\frac{16}{5}\)
\(\Rightarrow\frac{2}{5}\left[x+\frac{1}{3}\right]=1\)
\(\Rightarrow x+\frac{1}{3}=\frac{5}{2}\)
\(\Rightarrow x=\frac{13}{6}\)
DT
3
1 tháng 6 2020
Ta có:\(\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times..\times\frac{2018}{2019}\times\frac{2019}{2020}\)\(=\frac{1}{2020}\)
Vậy biểu thức \(\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times..\times\frac{2018}{2019}\times\frac{2019}{2020}\)\(=\frac{1}{2020}\)
1 tháng 6 2020
1/2 x 2/3 x 3/4 x ... x 2018/2019 x 2019/2020
= 1 x 2 x 3 x ... x 2018 x 2019 / 2 x 3 x 4 x ... x 2019 x 2020
Khử loại đi ta còn lại phân số 1/2020
Hok tốt ^^
KJ
2
4 tháng 7 2019
Số hạng tử của D là : ( 19 - 1 ) : 2 + 1 = 10
=> D = ( 19 + 1 ) . 10 : 2 = 100
Chúc bạn hok tốt !!
4 tháng 7 2019
Số phần tử của D là :
( 19 - 1 ) : 2 + 1 = 10
Vậy tổng của D là :
( 19 + 1 ) . 10 : 2 = 100
Hok tốt!!!!!!!!!!!!!!!!!!!!!!!
T
2 tháng 10 2018
\(a.\frac{19}{5}\cdot\frac{4}{7}+\frac{3}{7}\cdot\frac{19}{5}-\frac{4}{5}\)
\(=\frac{19}{5}\cdot\left(\frac{4}{7}+\frac{3}{7}\right)-\frac{4}{5}\)
\(=\frac{19}{5}\cdot1-\frac{4}{5}\)
\(=\frac{19}{5}-\frac{4}{5}=\frac{15}{5}=3\)
\(b.2\frac{2}{7}\cdot5\frac{2}{5}+\frac{16}{7}\cdot1\frac{3}{5}+\frac{1}{2}\)
\(=\frac{16}{7}\cdot\frac{27}{5}+\frac{16}{7}\cdot\frac{8}{5}+\frac{1}{2}\)
\(=\frac{16}{7}\cdot\left(\frac{27}{5}+\frac{8}{5}\right)+\frac{1}{2}\)
\(=\frac{16}{7}\cdot7+\frac{1}{2}\)
\(=16+\frac{1}{2}=\frac{33}{2}\)
\(c.\frac{3}{7}\cdot3\frac{3}{4}-\frac{3}{7}\cdot\frac{5}{4}-\frac{1}{4}\)
\(=\frac{3}{7}\cdot\frac{15}{4}-\frac{3}{7}\cdot\frac{5}{4}-\frac{1}{4}\)
\(=\frac{3}{7}\cdot\left(\frac{15}{4}-\frac{5}{4}\right)-\frac{1}{4}\)
\(=\frac{3}{7}\cdot\frac{5}{2}-\frac{1}{4}\)
\(=\frac{15}{14}-\frac{1}{4}=\frac{23}{28}\)
Chú ý: \(\cdot:\times\)
Mik làm luôn, ko vt lại đề
B = 1/1 - 1/3 + 1/3 - 1/5 +...+1/2017 - 1/2019
B = 1 - 1/2019
B = 2018/2019
TL:
\(B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2017}-\frac{1}{2019}\)
\(B=1-\frac{1}{2019}\)
\(B=\frac{2018}{2019}\)
Vậy \(B=\frac{2018}{2019}\)