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\(A=\frac{2016^{2016}+1}{2016^{2017}+1}\Rightarrow2016A=\frac{2016^{2017}+2016}{2016^{2017}+1}=1+\frac{2015}{2016^{2017}+1}\)
\(B=\frac{2016^{2017}-3}{2016^{2018}-3}\Rightarrow2016B=\frac{2016^{2018}-6048}{2016^{2018}-3}=1+\frac{-6045}{2016^{2018}-3}\)
Vì \(\frac{2015}{2016^{2017}+1}>0;\frac{-6045}{2016^{2018}-3}< 0\)
Nên: A>B
Ta có:
A = \(\dfrac{2017}{2019}=1-\dfrac{2}{2019}\)
B= \(\dfrac{2019}{2021}\) = 1- \(\dfrac{2}{2021}\)
Ta có:
\(\dfrac{2}{2019}>\dfrac{2}{2021}\)
=> 1- \(\dfrac{2}{2019}< 1-\dfrac{2}{2021}\)
=> \(\dfrac{2017}{2019}< \dfrac{2019}{2021}\)
Lại có \(\dfrac{1}{2}< \dfrac{2}{3}\)
=>\(\dfrac{2017}{2019}+\dfrac{1}{2}< \dfrac{2019}{2021}+\dfrac{2}{3}\)
Vậy A<B
\(Ta\)có :\(a\)=\(\frac{2017\cdot2018-1}{2017.2018}\)=\(\frac{2017.2018}{2017.2018}\)-\(\frac{1}{2017.2018}\)=1-\(\frac{1}{2017.2018}\)
\(b\)=\(\frac{2019.2020-1}{2019.2020}\)=\(\frac{2019.2020}{2019.2020}\)-\(\frac{1}{2019.2020}\)=1-\(\frac{1}{2019.2020}\)
Vì \(\frac{1}{2018.2019}\)> \(\frac{1}{2019.2020}\)nên \(a\)< \(b\)(sử dụng phần bù)
ta co
1/2.2<1/1*2
...
1/2018*2018<1/2017*2018
=>1/2*2+...+1/2018*1018<1/1*2+...+1/2017.2018
.....(tinh 1/1*2+...+1/2017.*2018)
=>1/2*2+...+1/2018*2018<1-1/2018<1
=>1/2*2+...+1/2018*2018<1
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 \(\frac{1991.1999}{1995.1995}< 1.\)
A = 2017 - ( 1 -\(\frac{3}{4}\)) - ( 1 - \(\frac{3}{5}\)) -...- ( 1 - \(\frac{3}{2020}\)) : Còn lại
A = 2017 - 1 + \(\frac{3}{4}\) - 1 + \(\frac{3}{5}\) -... - 1 +\(\frac{3}{2020}\) : Còn lại
A = 0 + \(\frac{3}{4}\)+\(\frac{3}{5}\)+\(\frac{3}{2020}\) : CÒn lại
A = 3 x ( \(\frac{1}{4}\)+\(\frac{1}{5}\)+...+ \(\frac{1}{2020}\)) : ( \(\frac{1}{4x5}\)+\(\frac{1}{5x5}\) + ... )
A = 3 : \(\frac{1}{5}\)
A = 15
= 15 nha
đúng 1000000000000000000000000000000% đó cái này ở violympic cấp tỉnh lớp 5 đó....
CÁC BẠN GIÚP MÌNH NHANH VỚI NHÉ! BẠN NÀO TRẢ LỜI ĐẦU TIÊN THÌ MÌNH SẼ K CHO!!!!!!!!!!
So sánh :a)20172018 và20182019 b)201,62017 và201,72018 c)20152019 và504505 d)2x2018+12x2019+1 và2x2017+12x2018+1 e)2x2018+12x2019+1 và3x2018+13x2019+1 f)223334 và22233334
* Xét số bị chia, ta có:
(2017 - 1) : 1 + 1 = 2017
(2020 - 4): 1 + 1 = 2017
Suy ra: Số hạng thứ hai của hiệu có số số hạng là: 2017
Suy ra: Ta có thể chia số 2017 thành 2017 số 1 để có:
2017 - 1/4 - 2/5 - 3/6 - 4/7 + …. - 2017/2020
= 1 - 1/4 + 1 - 2/5 + 1 - 3/6 + 1 - 4/7 + …. + 1 - 2017/2020
= 3/4 + 3/5 + 3/6 + 3/7 + …. + 3/2020 =
3 x (1/4 + 1/5 + 1/6 + 1/7 + …. 1/2020) (1)
* Xét số chia, ta có:
1/20 = 1/(4 x 5)
1/25 = 1/(5 x 5)
1/30 = 1/(6 x 5)
…
1/10100 = 1/(2020 x 5)
Suy ra:
1/20 + 1/25 + 1/30 + 1/35 + … + 1/10100
1/(4 x 5) + 1/25 + 1/30 + 1/35 + … + 1/(2020 x5 )
= 1/5 x (1/4 + 1/5 + 1/6 + 1/7 + …. + 1/2020) (2)
Ta thấy số bị chia (1) và số chia (2) có thừa số giống nhau là: (1/4 + 1/5 + 1/6 + 1/7 + …. 1/2020)
Suy ra: B = 3 : 1/5 = 15
