Bg (Sửa đề vì 1 + 3⁰ = 1 + 1 mà trong những bài toán ntn thì hai lần 1 hơi sai sai)

Ta có: A = 3⁰ + 3² + 3⁴ + 3⁶ +...+ 3²⁰²⁰ 

=> A = 3^2.0 + 3^2.1 + 3^2.2 + 3^2.3 +...+ 3^1010.2 

=> A = 3^2.0 + 3^2.1 + 3^2.2 + 3^2.3 +...+ 3^2.1010 

=> A = (3²)⁰ + (3²)¹ + (3²)² + (3²)³ +...+ (3²)¹⁰¹⁰ 

=> A = 9⁰ + 9¹ + 9² + 9³ +...+ 9¹⁰¹⁰ 

=> 9A = 9.(9⁰ + 9¹ + 9² + 9³ +...+ 9¹⁰¹⁰)

=> 9A = 9¹ + 9² + 9³ + 9⁴ +...+ 9¹⁰¹¹ 

=> 9A - A = (9¹ + 9² + 9³ + 9⁴ +...+ 9¹⁰¹¹) - (9⁰ + 9¹ + 9² + 9³ +...+ 9¹⁰¹⁰)

=> 8A = 9¹⁰¹¹  - 9⁰

=> 8A + 1 = 9¹⁰¹¹ + 1 - 1

=> 8A + 1 = 9¹⁰¹¹ 

=> 8A + 1 = (3²)¹⁰¹¹ 

=> 8A + 1 = 3ⁿ = 3²⁰²² 

Vậy n = 2022

a minh ko biet b minh ko lam duoc nhe . :)))))))

a,x=2 y=25

Từ \(\frac{2019a+2020c}{2019a-2021c}=\frac{2019b+2020d}{2019b-2021d}\)

=> \(\frac{2019a-2021c+4041c}{2019a-2021c}=\frac{2019b-2021d+4041d}{2019b-2021d}\)

=> \(1+\frac{4041c}{2019a-2021c}=1+\frac{4041d}{2019b-2021d}\)

=> \(\frac{4041c}{2019a-2021c}=\frac{4041d}{2019b-2021d}\)

=> \(4041c\left(2019b-2021d\right)=4041d\left(2019a-2021c\right)\)

=> \(c\left(2019b-2021d\right)=d\left(2019a-2021c\right)\)( rút 4041 ở cả hai vế )

=> \(2019bc-2021cd=2019ad-2021cd\)

=> \(2019ad-2021cd-2019bc+2021cd=0\)

=> \(2019\left(ad-bc\right)=0\)

=> \(ad-bc=0\)

=> \(ad=bc\)

=> \(\frac{a}{b}=\frac{c}{d}\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=kb\\c=kd\end{cases}}\)

+) \(\left(\frac{a+b}{c+d}\right)^{2020}=\left(\frac{kb+b}{kd+d}\right)^{2020}=\left[\frac{b\left(k+1\right)}{d\left(k+1\right)}\right]^{2020}=\left(\frac{b}{d}\right)^{2020}=\frac{b^{2020}}{d^{2020}}\)(1)

+) \(\frac{a^{2020}+b^{2020}}{c^{2020}+d^{2020}}=\frac{\left(kb\right)^{2020}+b^{2020}}{\left(kd\right)^{2020}+d^{2020}}=\frac{k^{2020}b^{2020}+b^{2020}}{k^{2020}d^{2020}+d^{2020}}=\frac{b^{2020}\left(k^{2020}+1\right)}{d^{2020}\left(k^{2020}+1\right)}=\frac{b^{2020}}{d^{2020}}\)(2)

Từ (1) và (2) ta có điều phải chứng minh

Tìm GTNN của biểu thức

A= 2a²+ab-2a+2024

Biết √[a+√(a²+2020)] . √[b+√(b²+2020)]=2020

Câu hỏi tương tự Đọc thêm Báo cáo

Toán lớp 9

Gọi d là ƯC( n + 2 ; 2n + 3 )

=> \(\hept{\begin{cases}n+2⋮d\\2n+3⋮d\end{cases}}\Rightarrow\hept{\begin{cases}2\left(n+2\right)⋮d\\2n+3⋮d\end{cases}}\Rightarrow\hept{\begin{cases}2n+4⋮d\\2n+3⋮d\end{cases}}\)

=> \(2n+4-\left(2n+3\right)⋮d\)

=> \(1⋮d\)=> \(d=1\)

=> ƯCLN( n + 2 ; 2n + 3 ) = 1

hay n + 2 ; 2n + 3 là hai số nguyên tố cùng nhau