![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\hept{\begin{cases}A=-\frac{1}{2020}-\frac{3}{2019^2}-\frac{5}{2019^3}-\frac{7}{2019^4}^{ }\\B=-\frac{1}{2020}-\frac{7}{2019^2}-\frac{5}{2019^3}-\frac{3}{2019^4}\end{cases}}\)
=>\(A-B=-\frac{1}{2020}-\frac{3}{2019^2}-\frac{5}{2019^3}-\frac{7}{2019^4}+\frac{1}{2020}+\frac{7}{2019^2}+\frac{5}{2019^3}+\frac{3}{2019^4}\)
\(=>A-B=\left(-\frac{3}{2019^2}+\frac{7}{2019^2}\right)+\left(-\frac{7}{2019^4}+\frac{3}{2019^4}\right)\)
=>\(A-B=\frac{4}{2019^2}+-\frac{4}{2019^4}\)
=>\(A-B=\frac{2019^2.4}{2019^4}-\frac{4}{2019^4}\)
=>\(A>B\)
cách này mình tự nghĩ
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(A=1-\dfrac{2\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}{4\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}\)
=1-2/4=1/2
b: \(B=\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)
\(=\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}=5\cdot\dfrac{-6}{9}=-\dfrac{10}{3}\)
c: x-y=0 nên x=y
\(C=x^{2020}-x^{2020}+y\cdot y^{2019}-y^{2019}\cdot y+2019\)
=2019
![](https://rs.olm.vn/images/avt/0.png?1311)
=>3B=1+1/3+1/3^2+...+1/3^2019
=>3B-B=(1+1/3+1/3^2+...+1/3^2019)-(1/3+1/3^2+1/3^3+...+1/3^2020)
<=>2B=1-1/3^2020= \(\frac{3^{2020}-1}{3^{2020}}\)
\(\Rightarrow B=\frac{3^{2020}-1}{3^{2020}.2}\)
#)Giải :
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2020}}\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2019}}\)
\(3B-B=2B=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2019}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2020}}\right)\)
\(2B=1-\frac{1}{3^{2020}}\)
\(B=\frac{1-\frac{1}{3^{2020}}}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\dfrac{3}{4}A=\dfrac{3}{4}-\left(\dfrac{3}{4}\right)^2+...+\left(\dfrac{3}{4}\right)^{2021}\)
=>\(\dfrac{7}{4}\cdot A=\left(\dfrac{3}{4}\right)^{2021}+1\)
=>\(A\cdot\dfrac{7}{4}=\dfrac{3^{2021}+4^{2021}}{4^{2021}}\)
=>\(A=\dfrac{3^{2021}+4^{2021}}{4^{2020}\cdot7}\)
b: Vì 3^2021+4^2021 ko chia hết cho 4^2020*7 nên A ko là số nguyên
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
\(A-B=\frac{4}{2019^2}-\frac{4}{2019^4}\)
Dễ thấy $0< 2019^2< 2019^4\Rightarrow \frac{4}{2019^2}> \frac{4}{2019^4}$
$\Rightarrow A-B=\frac{4}{2019^2}-\frac{4}{2019^4}>0$
$\Rightarrow A>B$
thầy ơi vì sao \(A-B=\frac{4}{2019^2}-\frac{4}{2019^4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\dfrac{x-2}{2020}-1+\dfrac{x-3}{2019}-1=\dfrac{x-2019}{3}-1+\dfrac{x-2020}{2}-1\)
=>x-2022=0
hay x=2022
Đặt A = \(3-3^2+3^3-3^4+...+3^{2019}-3^{2020}\)
3A = \(3^2-3^3+3^4-3^5+...+3^{2020}-3^{2021}\)
4A = \(3-3^{2021}\)
A = \(\frac{3-3^{2021}}{4}\)
Vậy .......
Hok tốt
Đặt \(A=3-3^2+...+3^{2019}-3^{2020}\)
\(3A=3^2-3^3+...+3^{2020}-3^{2021}\)
\(3A+A=\left(3^2+...-3^{2021}\right)+\left(3-3^2...-3^{2020}\right)\)
\(4A=3-3^{2021}\)
\(A=\frac{3-3^{2021}}{4}\)
hok tốt!!