Cho \(n=2^{2019}-2^{2018}-2^{2017}-...-2^2-2-1\) .Tính \(A=3^n+2^n+2020^n\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(N=2^{2019}-2^{2018}-2^{2017}-...-2-1\)
\(=2^{2019}-\left(2^{2018}+2^{2017}+...+2+1\right)\)
Đặt \(B=1+2+...+2^{2017}+2^{2018}\)
\(\Rightarrow\) \(2B=2+2^2+...+2^{2018}+2^{2019}\)
\(\Rightarrow\) \(B=2^{2019}-1\)
\(\Rightarrow\) \(N=2^{2019}-2^{2019}+1=1\)
\(\Rightarrow\) \(A=20^1+11^1+2019^1\)
\(=20+11+2019\)
\(=2050\)
Study well ! >_<
N=\(2^{2019}-\left(1+2+.....2^{2018}\right)\)
Đặt B=\(1+2+..........+2^{2018}\)
2B=\(2+2^2+..........+2^{2019}\)
2B-B=B=\(2^{2019}-1\)
Suy ra N=\(2^{2019}-2^{2019}+1=1\)
A=20+11+2019=2050
hok tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
n = \(2^{2020}-2^{2019}-2^{2018}-...-2-1\)
=> 2n = \(2^{2021}-2^{2020}-2^{2019}-2^{2018}-...-2^2-2\)
=> 2n - n = \(2^{2021}-2^{2020}-2^{2020}+1\)
=> \(n=2^{2021}-2.2^{2020}+1=1\)
=> \(A=2018.1-2019.1+2020.1=2019\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{3}{n-2018}+\frac{2}{n-2019}+\frac{1}{n-2020}=3\)
\(\Leftrightarrow\frac{3}{n-2018}-1+\frac{2}{n-2019}-1+\frac{1}{n-2020}-1=0\)
\(\Leftrightarrow\frac{3-\left(n-2018\right)}{n-2018}+\frac{2-\left(n-2019\right)}{n-2019}+\frac{1-\left(n-2020\right)}{n-2020}=0\)
\(\Leftrightarrow\frac{2021-n}{n-2018}+\frac{2021-n}{n-2019}+\frac{2021-n}{n-2020}=0\)
\(\Leftrightarrow\left(2021-n\right)\left(\frac{1}{n-2018}+\frac{1}{n-2019}+\frac{1}{n-2020}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2021-n=0\left(1\right)\\\frac{1}{n-2018}+\frac{1}{n-2019}+\frac{1}{n-2020}=0\left(2\right)\end{cases}}\)
Giải \(\left(1\right)\Leftrightarrow n=2021\).
Giải \(\left(2\right)\):
- Với \(n< 2018\)thì: \(\frac{1}{n-2018}< 0,\frac{1}{n-2019}< 0,\frac{1}{n-2020}< 0\)nên phương trình vô nghiệm.
- Với \(n=2018,n=2019,n=2020\)không thỏa điều kiện xác định.
- Với \(n>2020\)thì \(\frac{1}{n-2018}>0,\frac{1}{n-2019}>0,\frac{1}{n-2020}>0\) nên phương trình vô nghiệm.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n=2^{2019}-2^{2018}-...-2^1-1=2^{2019}-\left(2^{2018}+2^{2017}+...+2^1+1\right)\)
Đặt\(S=1+2+...+2^{2017}+2^{2018}\)
\(\Rightarrow2S=2+2^2+...+2^{2018}+2^{2019}\)
\(\Rightarrow2S-S=\left(2+2^2+...+2^{2018}+2^{2019}\right)-\left(1+2+...+2^{2017}+2^{2018}\right)\)
\(\Rightarrow S=2^{2019}-1\)
Mà\(n=2^{2019}-S\)
\(\Rightarrow n=2^{2019}-\left(2^{2019}-1\right)=1\)
\(\Rightarrow A=3^1+2^1+2020^1=2025\)
Happy new year :)))
Ta có : n = 22019 - 22018 - 22017 - .... - 22 - 2 - 1 (1)
=> 2n = 22020 - 22019 - 22018 - .... - 23 - 22 - 2 (2)
Lấy (2) trừ (1) theo vế ta có :
2n - n = (22020 - 22019 - 22018 - .... - 23 - 22 - 2) - (22019 - 22018 - 22017 - .... - 22 - 2 - 1)
=> n = 22020 - 22019 - 22019 + 1
=> n = 22020 - 2.22019 + 1 = 22020 - 22020 + 1 = 1
Khi đó A = 31 + 21 + 20201 = 3 + 2 + 2020 = 2025
Vậy A = 2025