3.5² + 15.7² - 8³ : 2

= 3.25 + 15.49 - 512 : 2

= 75 + 735 - 256

= 810 - 256

= 554

3.5² + 15.7² - 8³ : 2

= 3.25 + 15.49 - 512 : 2

= 75 + 735 - 256

= 810 - 256

= 554

\(a=1+3+3^2+..+3^{2006}\\ \Rightarrow3a=3+3^2+3^3+3^4+...+3^{2007}\\ \Rightarrow a-3a=1-3^{2007}\\ \Rightarrow-2a=1-3^{2007}\\ \Rightarrow a=\dfrac{-1-3^{2007}}{2}\)

Thay \(a,b\) vào biểu thức \(b-2a\), ta có:

\(3^{2007}+\dfrac{1-3^{2007}}{2}.2\\ =3^{2007}+1-3^{2007}\\ =\left(3^{2007}-3^{2007}\right)+1\\ =1.\)

Ta có:

Nhân 2 vế với 3

3A = 3 + 3² + ... + 3²⁰⁰⁷

3A - A= ( 3 + 3² +...+ 3²⁰⁰⁷ ) - ( 1 + 3 + 3² +...+ 3²⁰⁰⁶)

2A = 3²⁰⁰⁷ - 1

B - A = 3²⁰⁰⁷ - (3²⁰⁰⁷ + 1)

97x99=97x(100-1)=97x100-97=9700-97=9603 nhé

\(97.99\\ =97.\left(100-1\right)\\ =9700-97\\ =9603.\)

\(A=2^0+2^1+2^2+...+2^{100}\)

\(2\cdot A=2^1+2^2+2^3+...+2^{101}\)

\(A=2^{101}-1\)

Vậy \(A=B\)

siu

1/5

\(2A-A=\left(2^2+2^3+...+2^{21}\right)-\left(2+2^2+...+2^{20}\right)\)

\(A=2^{21}-2\)

B tương tự câu A

\(5C-C=\left(5^2+5^3+...+5^{51}\right)-\left(5+5^2+...+5^{50}\right)\)

\(C=\dfrac{5^{51}-5}{4}\)

\(3D-D=3+3^2+...+3^{101}-\left(1+3+...+3^{100}\right)\)

\(D=\dfrac{3^{101}-1}{2}\)

\(A=2^1+2^2+2^3+...+2^{20}\)

\(2\cdot A=2^2+2^3+2^4+...+2^{21}\)

\(A=2^{21}-2\)

\(B=2^1+2^3+2^5+...+2^{99}\)

\(4\cdot B=2^3+2^5+2^7+...+2^{101}\)

\(B=\)\(\left(2^{101}-2\right):3\)

\(C=5^1+5^2+5^3+...+5^{50}\)

\(5\cdot C=5^2+5^3+5^4+...+5^{51}\)

\(C=(5^{51}-5):4\)

\(D=3^0+3^1+3^2+...+3^{100}\)

\(3\cdot D=3^1+3^2+3^3+...+3^{101}\)

\(D=(3^{101}-1):2\)

a) n+6 chia hết cho n+2

n+2 chia hết cho n+2

nên (n+6)-(n+2) chia hết cho n+2

4 chia hết cho n-2

=> n-2 = 1;-1;2;-2;4;-4

=> n=3;1;4;0;6

d) n^2 +4 chia hết cho 4

n+1 chia hết cho n+1 nên (n+1)(n+1) chia hết cho n+1 hay n²+2n+1 chia hết cho n+1

=> (n^2+2n+1)-(n^2+4) chia hết cho n-1

=> 2n+1-4 chia hết cho n-1

=> 2n - 3 chia hết cho n-1

 n-1 chia hết cho n-1 nên 2n-2 chia hết cho n-1

=> (2n-2)-(2n-3) chia hết cho n-1

=> 1 chia hết cho n-1

=> n-1 = 1;-1

=> n=0 

nè bạn

\(\dfrac{10}{x-1}\) là số nguyên khi:

10 ⋮ \(x-1\)

\(\Rightarrow x-1\inƯ\left(10\right)=\left\{1;-1;2;-2;5;-5;10;-10\right\}\)

\(\Rightarrow x\in\left\{2;0;3;-1;6;-4;11;-9\right\}\)

Ta có:

\(A=2+2^2+2^3+...+2^{2024}\)

\(A=2\cdot1+2\cdot2+2^2\cdot2+...+2^{2023}\cdot2\)

\(A=2\cdot\left(1+2+2^2+...+2^{2023}\right)\)

Mà: \(1+2+2^2+...+2^{2023}\) nguyên

Nên A chia hết cho 2 vậy A là hợp số 

A=2+2 

2

 +2 

3

 +...+2 

2024

�

=

2

⋅

1

+

2

⋅

2

+

2

2

⋅

2

+

.

.

.

+

2

2023

⋅

2

A=2⋅1+2⋅2+2 

2

 ⋅2+...+2 

2023

 ⋅2

�

=

2

⋅

(

1

+

2

+

2

2

+

.

.

.

+

2

2023

)

A=2⋅(1+2+2 

2

 +...+2 

2023

 )

Mà: 

1

+

2

+

2

2

+

.

.

.

+

2

2023

1+2+2 

2

 +...+2 

2023

  nguyên

Nên A chia hết cho 2 vậy A là hợp số 

\(135⋮9;270⋮9;1818⋮9\Rightarrow A⋮9\)

\(9.2021⋮9;243⋮9\Rightarrow B⋮9\)

\(2601⋮9;99⋮9\Rightarrow C⋮9\)

\(9.1234⋮9;2021⋮̸9\Rightarrow D⋮̸9\)