4x+achia hết cho 2x+1
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\(A=1+3+3^2+..........+3^{11}\)
\(\Leftrightarrow A=\left(1+3\right)+\left(3^2+3^3\right)+.........+\left(3^{10}+3^{11}\right)\)
\(\Leftrightarrow A=1\left(1+3\right)+3^2\left(1+3\right)+.........+3^{10}\left(1+3\right)\)
\(\Leftrightarrow A=1.4+3^2.4+.......+3^{10}.4\)
\(\Leftrightarrow A=4\left(1+3^2+..........+3^{10}\right)⋮4\left(đpcm\right)\)
\(A=1+3^1+3^2+3^3+...+3^{2021}\\=(1+3^1)+(3^2+3^3)+(3^4+3^5)...+(3^{2020}+3^{2021})\\=4+3^2\cdot(1+3)+3^4\cdot(1+3)+...+3^{2020}\cdot(1+3)\\=4+3^2\cdot4+3^4\cdot4+...+3^{2020}\cdot4\\=4\cdot(1+3^2+3^4+...+3^{2020})\)
Vì \(4\cdot(1+3^2+3^4+...+3^{2020})\vdots4\)
nên \(A\vdots4\)
\(\text{#}Toru\)
thank you bạn character debate nha, ai vô trả lời thì cảm ơn nhiều!!
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 !!!
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\(A=n\left(n-1\right)\left(n+1\right)\left(n^2-4+5\right)\)
=> \(A=n\left(n+1\right)\left(n-1\right)\left(n^2-4\right)+n\left(n+1\right)\left(n-1\right)\cdot5\)
=> \(A=n\left(n+1\right)\left(n-1\right)\left(n-2\right)\left(n+2\right)+n\left(n-1\right)\left(n+1\right).5\)
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
(11^2n-2^6n)=121^n-64^n chắc chắn chia hết cho 121-64=57 (1)
vì n là số tự nhiên ko chia hết cho 5
suy ra n = 1;2;3;4;6...
suy ra n^4 - 1 chắc chắn chia hết cho 5 (2)
từ 1 va 2 ta co dpcm (CHO MÌNH CÁI ĐÚNG NHA)
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 :/
Đề bài là?