Chứng minh rằng
a) $8^{9}$- $2^{24}$ chia hết cho 14
b) $2^{2020}$ – ($2^{2019}$ + $2^{2018}$ +…+ $2^{1}$+ $2^{0}$ = 1
Nhanh nha 3h phải nộp rồi
Chứng minh rằng a) $8^{9}$- $2^{24}$ chia hết cho 14 b) $2^{2020}$ – ($2^{2019}$ + $2^{2018}$ +…+ $2^{1}$+ $2^{0}$ = 1 Nhanh nha 3h phải nộp rồi
By Savannah
Đáp án + Giải thích các bước giải:
`a)` `8^9-2^24=(2^3)^9-2^24=2^27-2^24=2^24(2^3-1)=2^23*2*7=2^23*14vdots14(đpcm)`
`b)` Đặt `A=2^0+2^1+…+2^2018+2^2019`
`=>2A=2^1+2^2+…+2^2019+2^2020`
`=>2A-A=(2^1+2^2+…+2^2019+2^2020)-(2^0+2^1+…+2^2018+2^2019)`
`=>A=2^2020-2^0=2^2020-1`
Do đó : `2^2020-A=2^2020-(2^2020-1)=2^2020-2^2020+1=1(đpcm)`
a) `8^9 – 2^24`
`= (2^3)^9 – 2^24`
`= 2^27 – 2^24`
`= 2^24 ( 2^3 -1)`
`= 2^24 . (8-1)`
`= 2^24 .7 vdots 7` và `2^24 .7 vdots 2`
Mà `(2;7)=1`
`=> 8^9 – 2^21 vdots 14`
Vậy `8^9 – 2^21 vdots 14`
b) `2^2020 – (2^0 + 2^1 + …+ 2^2018+ 2^2019)`
Đặt `A= 2^0 + 2^1 + …+ 2^2018+ 2^2019`
`2A= 2( 2^0 + 2^1 + …+ 2^2018+ 2^2019)`
`2A= 2 + 2^2 + …+ 2^2019 + 2^2020`
`2A -A = 2+ 2^2 + …+ 2^2019 + 2^2020 – 2^0 – 2^1 -…-2^2018 – 2^2019`
`A= 2^2020 – 2^0`
`A= 2^2020 -1`
`=> 2^2020 – (2^0 + 2^1 + …+ 2^2018) + 2^2019) = 2^2020 – (2^2020 -1)`
`= 2^2020 -2^2020 +1`
`= 1`
Vậy `2^2020 – (2^0 + 2^1 + …+ 2^2018 + 2^2019) =1`