Cho \(N=2^{2019}-2^{2018}-2^{2017}-...-2-1\)
Tính\(A=20^n+11^n+2019^n\)
Những câu hỏi liên quan
Cho \(n=2^{2019}-2^{2018}-2^{2017}-...-2^2-2-1\) .Tính \(A=3^n+2^n+2020^n\)
\(n=2^{2019}-2^{2018}-...-2^1-1=2^{2019}-\left(2^{2018}+2^{2017}+...+2^1+1\right)\)
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
(n+2017^2018).(n+2018^2019) chia hết cho 2
Tính :A= [(2018/1)+(2017/2)+(2016/3)+(2015/4)+...+(4/2015)+(3/2016)+(2/2017)+(1/2018)]/[(2019/1)+(2019/2)+(2019/3)+(2019/4)+...+(2019/2015)+(2019/2016)+(2019/2017)+(2019/2018)+(2019/2019)]
Tính :A= [(2018/1)+(2017/2)+(2016/3)+(2015/4)+...+(4/2015)+(3/2016)+(2/2017)+(1/2018)]/[(2019/2)+(2019/3)+(2019/4)+(2019/5)+...+(2019/2015)+(2019/2016)+(2019/2017)+(2019/2018)+(2019/2019)]
2019+ 2018 + 2017 +.... + x = 2019 tìm số nguyên n thỏa mãn : 1! + 2! + 3! + .... + n! là một số chính phương
Cho M=2018 +20182+20183+...+20182018
CMR M chia het cho 2019
S2M Voi N=22019/2017 tat ca - 1
\(M=\left(2018+2018^2\right)+\left(2018^3+2018^4\right)+...+\left(2018^{2017}+2018^{2018}\right)\)
\(=2018\left(1+2018\right)+2018^3\left(1+2018\right)+...+2018^{2017}\left(1+2018\right)\)
\(=2018.2019+2018^3.2019+...+2018^{2017}.2019\)
\(=2019\left(2018+2018^3+...+2018^{2017}\right)⋮2019\)
b/ \(M=2018+2018^2+...+2018^{2018}\)
\(2018M=2018^2+2018^3+...+2018^{2018}+2018^{2019}\)
Lấy dưới trừ trên:
\(2018M-M=-2018+2018^{2019}\)
\(\Rightarrow2017M=2018^{2019}-2018\)
\(\Rightarrow M=\frac{2018^{2019}-2018}{2017}=\frac{2018^{2019}}{2017}-\frac{2017+1}{2017}=\frac{2018^{2019}}{2017}-1-\frac{1}{2017}\)
\(\Rightarrow M=N-\frac{1}{2017}\Rightarrow M< N\)
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Câu 1:Tìm tất cả số nguyên dương n sao cho tât cả số n+1,n+5,n+7,n+13,+n+17,n+25,n+37 đều là số nguyên tốCâu 2:Cho A1-frac{1}{2}+frac{1}{3}+frac{1}{4}+...+frac{1}{2017}-frac{1}{2018}+frac{1}{2019} và Bfrac{1}{1010}+frac{1}{1011}+...+frac{1}{2018}+frac{1}{2019}.Tính left(A-B-1right)^{2019}
Đọc tiếp
Câu 1:Tìm tất cả số nguyên dương n sao cho tât cả số n+1,n+5,n+7,n+13,+n+17,n+25,n+37 đều là số nguyên tố
Câu 2:Cho A=\(1-\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}\)
và B=\(\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018}+\frac{1}{2019}\).Tính \(\left(A-B-1\right)^{2019}\)
1. \(n\in\left\{1;2;3;4;5;...\right\}\)
2. \(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2019}\)
\(\Rightarrow A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}+\frac{1}{2019}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(\Rightarrow A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{1009}\)
\(\Rightarrow A=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2019}\)
Ta có :
\(\left(A-B-1\right)^{2019}=\left(\frac{1}{1010}+...+\frac{1}{2019}-\left(\frac{1}{1010}+...+\frac{1}{2019}\right)-1\right)^{2019}\)
\(=\left(-1\right)^{2019}=-1\)
\(\left(|x|-2017\right)^{\left(n+2018\right)\cdot\left(n+2019\right)}=-\left(2^3-3^2\right)^{2019}\)
\(\left(\left|x\right|-2017\right)^{\left(n+2018\right)\left(n+2019\right)}=-\left(2^3-3^2\right)^{2019}\)
\(\left(\left|x\right|-2017\right)^{\left(n+2018\right)\left(n+2019\right)}=-\left(-1\right)^{2019}=1\)
\(\Rightarrow\orbr{\begin{cases}\left(n+2018\right)\left(n+2019\right)=0\\\left|x\right|-2017=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}\orbr{\begin{cases}n=-2018\\n=-2019\end{cases}}\\\orbr{\begin{cases}x=2018\\x=-2018\end{cases}}\end{cases}}\)
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1. Tính giá trị biểu thức
a,A=1*2*3*...*2018*2019 - 1*2*3*...*2017*2018 - *1*2*3*...*2017*20182
b,B=(150-1/9-2/10-3/11-...-150/158):(1/36+1/40+1/44+....+1/632)
2, Chứng minh rằng phân số 4n+1/5n+1 là phân số tối giản với mọi số nguyên n
trình bày
Câu 1
a) A=2018!.(2019 - 1 -2018)
=2018!.0
= 0
vậy A= 0
b)\(B=\left(1-\frac{1}{9}+1-\frac{2}{10}+1+\frac{3}{11}+...+1-\frac{150}{158}\right):\left(\frac{1}{4}.\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{158}\right)\right)\)
\(=\left(\frac{8}{9}+\frac{8}{10}+...+\frac{8}{158}\right):\left(\frac{1}{4}\left(\frac{1}{9}+\frac{1}{10}+...+\frac{1}{158}\right)\right)\)
\(=8.\left(\frac{1}{9}+\frac{1}{10}+...+\frac{1}{158}\right):\left(\frac{1}{4}\left(\frac{1}{9}+\frac{1}{10}+...+\frac{1}{158}\right)\right)\)
\(=8:\frac{1}{4}\)
=32
Vậy B= 32
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