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Ta có:
\(10A=\frac{10^{2015}+20200}{10^{2015}+2020}=1+\frac{18180}{10^{2015}+2020}\)
\(10B=\frac{10^{2016}+20200}{10^{2016}+2020}=1+\frac{18180}{10^{2016}+2020}\)
Vì \(10^{2016}+2020>2^{2015}+2020\)
=> \(\frac{18180}{10^{2016}+2020}< \frac{18180}{10^{2015}+2020}\)
=> \(1+\frac{18180}{10^{2016}+2020}< 1+\frac{18180}{10^{2015}+2020}\)
=> 10B < 10A
=> B<A
Ta có :
B = \(\dfrac{2015}{1}+\dfrac{2014}{2}+\dfrac{2013}{3}+...+\dfrac{2}{2014}+\dfrac{1}{2015}\) => B = \(\left(1+\dfrac{2014}{2}\right)+\left(1+\dfrac{2013}{3}\right)+...+\left(1+\dfrac{2}{2014}\right)+\left(1+\dfrac{1}{2015}\right)+1\) => B = \(\dfrac{2016}{2}+\dfrac{2016}{3}+...+\dfrac{2016}{2014}+\dfrac{2016}{2015}+\dfrac{2016}{2016}\) => B = \(2016\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}\right)\) Ta có :
\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}}{2016\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}\right)}\)
=> \(\dfrac{A}{B}=\dfrac{1}{2016}\)
Vậy \(\dfrac{A}{B}=\dfrac{1}{2016}\)
Ta có: \(B=\dfrac{2017+2018+2019}{2018+2019+2020}=\dfrac{2017}{2018+2019+2020}+\dfrac{2018}{2018+2019+2020}+\dfrac{2019}{2018+2019+2020}\)
Mà \(\dfrac{2017}{2018}>\dfrac{2017}{2018+2019+2020}\)
\(\dfrac{2018}{2019}>\dfrac{2018}{2018+2019+2020}\)
\(\dfrac{2019}{2020}>\dfrac{2019}{2018+2019+2020}\)
\(\Rightarrow\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}>\dfrac{2017}{2018+2019+2020}+\dfrac{2018}{2018+2019+2020}+\dfrac{2019}{2018+2919+2020}\)
\(\Rightarrow A>B.\)
Vậy \(A>B.\)
Ta có: \(A=\frac{19^5+2016}{19^5-1}=\frac{19^5-1+2017}{19^5-1}=\frac{19^5-1}{19^5-1}+\frac{2017}{19^5-1}=1+\frac{2017}{19^5-1}\)
\(B=\frac{19^5+2015}{19^5-2}=\frac{19^5-2+2017}{19^5-2}=\frac{19^5-2}{19^5-2}+\frac{2017}{19^5-2}=1+\frac{2017}{19^5-2}\)
Vì \(\frac{2017}{19^5-1}< \frac{2017}{19^5-2}\Rightarrow1+\frac{2017}{19^5-1}< 1+\frac{2017}{19^5-2}\Rightarrow A< B\)
Vậy A < B
\(A=\frac{19^5+2016}{19^5-1}=\frac{\left(19^5-1\right)+2017}{19^5-1}=1+\frac{2017}{19^5-1}\)
\(B=\frac{19^5+2015}{19^5-2}=\frac{\left(19^5-2\right)+2017}{19^5-2}=1+\frac{2017}{19^5-2}\)
Vì \(19^5-1>19^5-2\) nên \(\frac{2017}{19^5-1}< \frac{2}{19^5-2}\)
\(\Rightarrow1+\frac{2017}{19^5-1}< 1+\frac{2017}{19^5-2}\)
Vậy \(A< B\)
Không tính thì sao mà làm được :)
a)
\(2020-\dfrac{1}{3^2}-\dfrac{1}{4^2}-...-\dfrac{1}{2019^2}\)
\(=3+\left(1-\dfrac{1}{3^2}\right)+\left(1-\dfrac{1}{4^2}\right)+....+\left(1-\dfrac{1}{2019^2}\right)\)
\(=3+\left(\dfrac{3^2-1}{3^2}+\dfrac{4^2-1}{4^2}+...+\dfrac{2019^2-1}{2019^2}\right)\)
\(=3+\left(\dfrac{2\cdot4}{3^2}+\dfrac{3\cdot5}{4^2}+\dfrac{4\cdot6}{5^2}+\dfrac{5\cdot7}{6^2}+...+\dfrac{2018\cdot2020}{2019^2}\right)\)
\(=3+\dfrac{\left(2\cdot3\cdot4\cdot....\cdot2018\right)}{3\cdot4\cdot5\cdot6...\cdot2019}\cdot\dfrac{\left(3\cdot4\cdot5\cdot....\cdot2020\right)}{3\cdot4\cdot5\cdot6\cdot....\cdot2019}=3+\dfrac{2\cdot2020}{2019}\)
\(=\dfrac{10097}{2019}\)
Có: \(\dfrac{1}{k^2}=\dfrac{1}{k.k}< \dfrac{1}{\left(k-1\right)k}\left(k\in\text{ℕ},k>0\right)\)
\(\Rightarrow A=2020-\dfrac{1}{3^2}-\dfrac{1}{4^2}-\dfrac{1}{5^2}-...-\dfrac{1}{2019^2}\)
\(A=2020-\left(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{2019^2}\right)\)
\(>2020-\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2018.2019}\right)\)
Có: \(\dfrac{1}{k-1}-\dfrac{1}{k}=\dfrac{1}{k\left(k-1\right)}\left(k\in\text{ℕ},k>0\right)\)
\(\Rightarrow A>2020-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-...+\dfrac{1}{2018}-\dfrac{1}{2019}\right)\)
\(A>2020-\dfrac{1}{2}+\dfrac{1}{2019}\)>2,2
Có: \(B=\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{17}\)
\(B=\dfrac{1}{5}+\left(\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}\right)\)\(< \dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{6}+...+\dfrac{1}{6}\)
\(=\dfrac{1}{5}+\dfrac{1}{6}.12=2+\dfrac{1}{5}=2,2\)
Vậy A>B.
Ta có:
n+2016/n+2019
=n+2015+1/n+2019
=(n+2015/n+2019)+(1/n+2019)
Vì n+2015/n+2019>n+2015/n+2020
=>n+2016/n+2019>n+2015/n+2020
a: \(\dfrac{-22}{45}=\dfrac{-22\cdot103}{45\cdot103}=\dfrac{-2266}{45\cdot103}\)
\(\dfrac{-51}{103}=\dfrac{-51\cdot45}{45\cdot103}=-\dfrac{2295}{45\cdot103}\)
mà -2266>-2295
nên \(-\dfrac{22}{45}>-\dfrac{51}{103}\)
b: Đặt \(A=\dfrac{2020^{2014}+1}{2020^{2015}+1};B=\dfrac{2020^{2015}-2}{2020^{2016}-2}\)
\(2020A=\dfrac{2020^{2015}+2020}{2020^{2015}+1}=1+\dfrac{2019}{2020^{2015}+1}\)
\(2020B=\dfrac{2020^{2016}-4040}{2020^{2016}-2}=1-\dfrac{4038}{2020^{2016}-2}\)
mà \(\dfrac{2019}{2020^{2015}+1}>\dfrac{-4038}{2020^{2016}-2}\)
nên 2020A>2020B
=>A>B