giúp me plssss

7 chia het cho (2x+1)

ma 7 chia het cho 1;7

=>2x+1=1=>x=0

2x+1=7=>x=3

ket luan x = 0;3

từ từ thôi cái này tốn có 4 câu hỏi thôi mà cho vào  1 câu làm gì

Lời giải:
Đặt $A=1-2+2^2-2^3+2^4-2^5+2^6-....-2^{2021}+2^{2022}$

$A=1+(-2+2^2-2^3)+(2^4-2^5+2^6)+(-2^7+2^8-2^9)+...+(2^{2020}-2^{2021}+2^{2022})$

$A=1+(-2+2^2-2^3)+2^3(2-2^2+2^3)+2^6(-2+2^2-2^3)+....+2^{2019}(2-2^2+2^3)$

$=1+(-6)+2^3.6+2^6(-6)+....+2^{2019}.6$

$=1+6(-1+2^3-2^6+...+2^{2019})$

Suy ra $A$ chia $6$ dư $1$/

1.

3²⁰¹⁵ = 3²⁰¹².3³ = (3⁴)⁵⁰³.27 = ...........1.27 = ..........7

35³² = (35⁴)⁸ = ........5

=> 3²⁰¹⁵ - 35³² = ................7 - ....................5 = ..................2 chia hết cho 2

a,  (3n+2) - (n-6) = 3n+2-n+6 = 2n+8 luôn chia hết cho 2

b, (n+2) + (n+4) + 6 = n+2+n+4+6 = 2n+12 luôn chia hết cho 2

c, (n+3)+2(n+4)+1 = n+3+2n+8+1 = 3n+12 luôn chia hết cho 3

Đặt \(B=4^{2023}+4^{2022}+\cdots+4^2+5\)

=>\(B=4^{2023}+4^{2022}+\cdots+4^2+4+1\)

=>\(4B=4^{2024}+4^{2023}+\cdots+4^3+4^2+4\)

=>\(4B-B=4^{2024}+4^{2023}+\cdots+4^3+4^2+4-4^{2023}-4^{2022}-\cdots-4^2-4-1\)

=>\(3B=4^{2024}-1\)

Ta có: \(A=75\cdot\left(4^{2023}+4^{2022}+\cdots+4^2+5\right)+25\)

\(=3\cdot25\cdot\left(4^{2023}+4^{2022}+\cdots+4^2+4+1\right)+25\)

\(=25\cdot3B+25=25\left(3B+1\right)=25\cdot4^{2024}\) ⋮\(4^{2024}\)