1.

\(B=1+3^1+....+3^{99}\\ \Rightarrow3.B=3+3^2+...+3^{100}\\ \Rightarrow2B=3^{100}-1\\ \Rightarrow B=\dfrac{3^{100}-1}{2}\)

\(\Rightarrow A=3^{100}-B=3^{100}-\dfrac{3^{100}-1}{2}\)

3.

a;b là số nguyên tố lớn hơn 3

=> a;b không chia hết cho 3 và a;b lẻ

a;b không chia hết cho 3 => a^2 ; b^2 chia 3 dư 1

=> A chia hết 3

TT : A chia hết 8

(3;8)=1 => A chia hết 24

1) \(3^x+3^{x+1}+3^{x+2}=351\)

\(\Rightarrow3^x\left(1+3^1+3^2\right)=351\)

\(\Rightarrow3^x.13=351\)

\(\Rightarrow3^x=27\)

\(\Rightarrow3^x=3^3\)

\(\Rightarrow x=3\)

2) \(C=2+2^2+2^3+2^4+...+2^{97}+2^{98}+2^{99}+2^{100}\)

\(\Rightarrow C=\left(2+2^2+2^3+2^4\right)+2^4\left(2+2^2+2^3+2^4\right)...+2^{96}\left(2+2^2+2^3+2^4\right)\)

\(\Rightarrow C=30+2^4.30...+2^{96}.30\)

\(\Rightarrow C=\left(1+2^4+...+2^{96}\right).30⋮30\)

mà \(30=5.6\)

\(\Rightarrow C⋮5\left(dpcm\right)\)

1,

Có \(3^x\)+ \(3^{x+1}\) + \(3^{x+2}\) = \(351\)

=> \(3^x\) + \(3^x\).\(3\) + \(3^x\).\(9\) = \(351\)

=> \(3^x\).\(13\) = \(351\)

=> \(3^x\) = \(27\)

=> \(x\) = \(3\)

2,

C = \(2\) + \(2^2\) + \(2^3\) + ... + \(2^{100}\)

2C = \(2^2\) + \(2^3\) + \(2^4\) + ... + \(2^{101}\)

2C - C = \(2^{101}\) - \(2\)

C = \(2^{101}\) - \(2\)

C = \(2\).\(\left(2^{100}-1\right)\)

C = 2.\(\left(\left(2^5\right)^{20}-1^{20}\right)\)

Có \(2^5\) \(-1\) \(⋮\) 5

=> \(\left(\left(2^5\right)^{20}-1^{20}\right)\) \(⋮\) 5

=> C \(⋮\) 5

3,

Xét \(\overline{abcdeg}\)

= \(\overline{ab}\).\(10000\) + \(\overline{cd}\).\(100\) + \(\overline{eg}\)

= \(\left(\overline{ab}+\overline{cd}+\overline{eg}\right)\) + \(9.\left(1111.\overline{ab}+11.\overline{cd}\right)\)

Có\(\left\{{}\begin{matrix}9.\left(1111.\overline{ab}+11.\overline{cd}\right)⋮9\left(1111.\overline{ab}+11.\overline{cd}\inℕ^∗\right)\\\overline{ab}+\overline{cd}+\overline{eg}⋮9\end{matrix}\right.\)

=> \(\overline{abcdeg}⋮9\)

4,

S = \(3^0+3^2+3^4+...+3^{2002}\)

9S = \(3^2+3^4+3^6+...+3^{2004}\)

9S - S = \(3^2+3^4+3^6+...+3^{2004}\) - (\(3^0+3^2+3^4+...+3^{2002}\))

8S = \(3^{2004}-1\)

=> 8S \(< 3^{2004}\)

b1

ta có : n+4 = (n+1)+3

=>n+1+3 chia hết cho n+1

vì n+1 chia hết cho n+1

=>3 chia hết cho n+1

=> n+1 chia hết cho 3

=> n+1 thuộc Ư 3 =[1;3]

=> n+1=1                   n+1=3

     n    =1-1                n    =3-1

     n    =0                   n    =2

vậy n thuộc [0;2]

S=1-3+3\(^2\)-....+3\(^{98}\)-3\(^{99}\)(1)

\(\Rightarrow\)3S=3-3\(^2\)+3\(^3\)+...+3\(^{99}\)-3\(^{100}\)(2)

Từ(1)và(2)\(\Rightarrow\)4S=1-3\(^{100}\)

Do S chia hết cho -20\(\Rightarrow\)4S chia hết cho -20

\(\Rightarrow\)4S chia hết cho 4\(\Rightarrow\)1-3\(^{100}\)chia hết cho 4

\(\Rightarrow\)3\(^{100}\)chia hết 4 dư 1

Bài 1:

\(M\left(1\right)=a+b+6\)

Mà \(M\left(1\right)=0\)

\(\Rightarrow a+b+6=0\)

\(\Rightarrow a+b=-6\)( * )

\(\Rightarrow2a+2b=-12\) (1)

Ta có: \(M\left(-2\right)=4a-2b+6\)

Mà \(M\left(-2\right)=0\)

\(\Rightarrow4a-2b=-6\)(2)

Lấy (1) cộng (2) ta được:

\(6a=-18\)

\(a=-3\)

Thay a=-3 vào (* ) ta được:

\(b=-3\)

Vậy a=-3 ; b=-3

Bài 2:

a) \(\frac{5}{x}+\frac{y}{4}=\frac{1}{8}\)

\(\Leftrightarrow\frac{1}{8}-\frac{y}{4}=\frac{5}{x}\)

\(\Leftrightarrow\frac{1}{8}-\frac{2y}{8}=\frac{5}{x}\)

\(\Leftrightarrow\frac{1-2y}{8}=\frac{5}{x}\)

\(\Leftrightarrow\left(1-2y\right).x=5.8\)

\(\Leftrightarrow\left(1-2y\right).x=40\)

Vì \(x,y\in Z\Rightarrow1-2y\in Z\)

mà \(40=1.40=40.1=5.8=8.5=\left(-1\right).\left(-40\right)=\left(-40\right).\left(-1\right)=\left(-5\right).\left(-8\right)=\left(-8\right).\left(-5\right)\)

Thử từng TH

Bài 4: 

b) Ta có: \(x=99\Rightarrow100=x+1\)

Ta có: \(P\left(99\right)=x^{99}-\left(x+1\right).x^{98}+\left(x+1\right)x^{97}-...+\left(x+1\right)x-1\)

\(=x^{99}-x^{99}-x^{98}+x^{98}+x^{97}-...+x^2+x-1\)

\(=x+1\)(1)

Thay x=99 vào (1) ta được:

\(P\left(99\right)=99+1\)

\(=100\)