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8 tháng 2 2022

x là 0, số 0 chia hết cho tất cả các số tự nhiên N*. và 4x+a=0,  2x +1=1,

0:1=0 :/

9 tháng 1 2022

Đề bài là?

4 tháng 11 2017

\(A=1+3+3^2+..........+3^{11}\)

\(\Leftrightarrow A=\left(1+3\right)+\left(3^2+3^3\right)+.........+\left(3^{10}+3^{11}\right)\)

\(\Leftrightarrow A=1\left(1+3\right)+3^2\left(1+3\right)+.........+3^{10}\left(1+3\right)\)

\(\Leftrightarrow A=1.4+3^2.4+.......+3^{10}.4\)

\(\Leftrightarrow A=4\left(1+3^2+..........+3^{10}\right)⋮4\left(đpcm\right)\)

4 tháng 11 2017

A = 1 + 3 + 32 + 33 + ... + 311

A = ( 1 + 3 ) + ( 32 + 33 ) + ... + ( 310 + 311 )

A = 4 + 32 . ( 1 + 3 ) + ... + 310 . ( 1 + 3 )

A = 4 + 32 . 4 + ... + 310 . 4

A = 4 . ( 1 + 32 + ... + 310 ) \(⋮\) 4 ( Vì trong tích có một thừa số chia hết cho 4 )

~ Chúc bạn học giỏi ! ~

5 tháng 11 2023

\(A=1+3^1+3^2+3^3+...+3^{2021}\\=(1+3^1)+(3^2+3^3)+(3^4+3^5)...+(3^{2020}+3^{2021})\\=4+3^2\cdot(1+3)+3^4\cdot(1+3)+...+3^{2020}\cdot(1+3)\\=4+3^2\cdot4+3^4\cdot4+...+3^{2020}\cdot4\\=4\cdot(1+3^2+3^4+...+3^{2020})\)

Vì \(4\cdot(1+3^2+3^4+...+3^{2020})\vdots4\)

nên \(A\vdots4\)

\(\text{#}Toru\)

5 tháng 11 2023

thank you bạn character debate nha, ai vô trả lời thì cảm ơn nhiều!!

 

23 tháng 10 2023

a) A = 4 + 4² + 4³ + ... + 4¹²

= 4.(1 + 4 + 4² + 4³ + ... + 4¹¹) ⋮ 4

Vậy A ⋮ 4

b) A = 4 + 4² + 4³ + 4⁴ + ... + 4¹²

= (4 + 4²) + (4³ + 4⁴) + ... + (4¹¹ + 4¹²)

= 4.(1 + 4) + 4³.(1 + 4) + ... + 4¹¹.(1 + 4)

= 4.5 + 4³.5 + ... + 4¹¹.5

= 5.(4 + 4³ + ... + 4¹¹) ⋮ 5

Vậy A ⋮ 5

c) A = 4 + 4² + 4³ + 4⁴ + ... + 4¹²

= (4 + 4² + 4³) + (4⁴ + 4⁵ + 4⁶) + ... + (4¹⁰ + 4¹¹ + 4¹²)

= 4.(1 + 4 + 4²) + 4⁴.(1 + 4 + 4²) + ... + 4¹⁰.(1 + 4 + 4²)

= 4.21 + 4⁴.21 + ... + 4¹⁰.21

= 21.(4 + 4⁴ + ... + 4¹⁰) ⋮ 21

Vậy A ⋮ 21

1) ta có : 180= 9 x 5 x 4

Mà 1494 chia hết cho 9; 1495 chia hết cho 5 và 1496 chia hết cho 4.

=> 1494 x 1495 x 1496 chia hết cho 180.

2)Ta có 495= 9 x 5 x 11

Mà 1494 chia hết cho 9; 1495 chia hết cho 5 và 1496 chia hết cho 11.

=> 1494 x 1495 x 1496 chia hết cho 495.

k mình nha !!!

16 tháng 10 2022

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4 tháng 4 2018

\(A=n\left(n-1\right)\left(n+1\right)\left(n^2-4+5\right)\)

=> \(A=n\left(n+1\right)\left(n-1\right)\left(n^2-4\right)+n\left(n+1\right)\left(n-1\right)\cdot5\)

=> \(A=n\left(n+1\right)\left(n-1\right)\left(n-2\right)\left(n+2\right)+n\left(n-1\right)\left(n+1\right).5\)

Vì \(n\left(n+1\right)\left(n+2\right)\left(n-1\right)\left(n-2\right)\)là tích 5 số nguyên liên tiếp nên chia hết cho 2 và 5 => chia hết cho 2.5=10 

n(n+1)(n-1) là tích 3 số nguyên liên tiếp nên chia hết cho 2 => 5n(n+1)(n-1) chia hết cho 2.5=10 

=> A chia hết cho 10 đpcm