9^2 + 9^3 + .... + 9^2022
Help me!
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
( 1/2 + 1/3 + 1/4 + ... + 1/10 ) . x = 1/9 + 2/8 + ... + 9/1
=> x = ( 1/9 + 2/8 + ... + 9/1 ) : ( 1/2 + 1/3 + ... 1/10 )
=> x = ( 9/1 + 8/2 + ... + 2/8 + 1/9 ) : ( 1/2 + 1/3 + 1/4 + ... + 1/10 )
=>x = [ ( 9 - 1 - 1 -... - 1 ) +( 8/2 + 1 ) + ( 7/3 + 1 ) + ... + ( 1/9 + 1 ) ] : ( 1/2 + 1/3 + 1/4 + ... + 1/10 )
=> x = ( 1 + 10/2 + 10/3 + ... + 10/9 ) : ( 1/2 + 1/3 + 1/4 + ... + 1/10 )
=> x = [10 . ( 1/2 + 1/3 + ... + 1/9 ) ] : ( 1/2 + 1/3 + 1/4 + ... + 1/10 )
=> x = 10
Chúc Bạn Học Tốt
#𝗝𝘂𝗻𝗻
\(=\dfrac{3^9\cdot3^2}{3^9}\cdot2022=3^2\cdot2022=9\cdot2022=18198\)
a) \(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}-\dfrac{3}{4}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{72}\)
\(=\dfrac{5+9+1}{15}-\dfrac{27+8+1}{36}+\dfrac{1}{72}=1-1+\dfrac{1}{72}=\dfrac{1}{72}\)
b) \(=\dfrac{1}{5}-\dfrac{1}{5}-\dfrac{3}{7}+\dfrac{3}{7}+\dfrac{5}{9}-\dfrac{5}{9}-\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{7}{13}-\dfrac{7}{13}-\dfrac{9}{16}\)
\(=\dfrac{9}{16}\)
\(\dfrac{3}{5}+\dfrac{5}{7}\) x \(\dfrac{3}{5}-\dfrac{9}{7}\) x \(\dfrac{3}{5}\)
\(=\dfrac{3}{5}+\left(\dfrac{5}{7}-\dfrac{9}{7}\right)\) x \(\dfrac{3}{5}\)
\(=\dfrac{3}{5}-\dfrac{4}{7}\) x \(\dfrac{3}{5}\)
\(=\dfrac{3}{5}-\dfrac{12}{35}\)
\(=\dfrac{21}{35}-\dfrac{12}{35}=\dfrac{21-12}{35}=\dfrac{9}{35}\)
A=1−2−3+4−5−6+7−8−9+....+2020−2021−2022D=1-2-3+4-5-6+7-8-9+....+2020-2021-2022
A =(1−2−3)+(4−5−6)+(7−8−9)+....+(2020−2021−2022)D=(1-2-3)+(4-5-6)+(7-8-9)+....+(2020-2021-2022)
A=(−4)+(−7)+(−10)+.....+(−2023)D=(-4)+(-7)+(-10)+.....+(-2023)
A=[(2023−4):3+1].[(−2023−4):2]D=[(2023-4):3+1].[(-2023-4):2]
A=674.(−1013,5)D=674.(-1013,5)
A=−683099
A=1−2−3+4−5−6+7−8−9+....+2020−2021−2022D=1-2-3+4-5-6+7-8-9+....+2020-2021-2022
A =(1−2−3)+(4−5−6)+(7−8−9)+....+(2020−2021−2022)D=(1-2-3)+(4-5-6)+(7-8-9)+....+(2020-2021-2022)
A=(−4)+(−7)+(−10)+.....+(−2023)D=(-4)+(-7)+(-10)+.....+(-2023)
A=[(2023−4):3+1].[(−2023−4):2]D=[(2023-4):3+1].[(-2023-4):2]
A=674.(−1013,5)D=674.(-1013,5)
A=−683099
Đặt `A=9^2+9^3+...+9^2021+9^2022`
`=> 9A=9^3+9^4+...+9^2022+9^2023`
`=> 9A-A=(9^3+9^4+...+9^2022+9^2023)-(9^2+9^3+...+9^2021+9^2022)`
`=> 8A=9^2023-9`
`=> A=(9^2023-9)/8`
888888888888X9999999999999