a, n=-2

b,n-2 thuoc u cua 5 

XXX

BEEG XEX XXEX = PHIM XEX

Ai giúp nguyen phan thu ngan

thì hãy cho mình 1 k

cảm ơn các bjan nhìu!!

chỉ tăng có 1 cái thôi thì chán l.tăng ken hắn 5 cái đi

có thể đ là 1 điều có khả thi . mik sẽ gi h nếu có 5 cái k

a ) do n+5 chia het  cho n-2 => (n-2)+7 chia het cho n-2

ma n-2 chia het cho n-2

=>7 chia het cho n-2 

=> n-2 thuoc {1;2;-1;-2}

=> n thuoc {3;4;1;0}

b) do n-1 chia het cho n-1 => 2.(n-1) chia het cho n-1=> 2n- 2 chia het cho n-1

ma 2n chia het cho n-1

=>-2 chia het cho n-1

=>n-1 thuoc {1;2;-1;-2}

=>n thuoc {2;3;0;-1

a) n+5 chia hết cho n-2

=>n-2+7 chia hết cho n-2

Mà n-2 chia hết cho n-2

=>7 chia hết cho n-2

=>n-2\(\in\)Ư(7)

=>n-2\(\in\){-7;-1;1;7}

=>n\(\in\){-5;1;3;9}

b) 2n chia hết cho n-1

=>n+n chia hết cho n-1

=>n-1+n-1+2 chia hết cho n-1

=>2(n-1)+2 chia hết cho n-1

Mà 2(n-1) chia hết cho n-1

=>2 chia hết cho n-1

=>n-1\(\in\){-2;-1;1;2}

=>n\(\in\){-1;0;2;3}

\(n^2+3\)chia hết cho n - 1

\(\Rightarrow\)\(n^2-1+4\) chia hết cho n - 1

\(\Rightarrow\)\(n^2-1^2+4\) chia hết cho n - 1

(n - 1)(n + 1) + 4 chia hết cho n - 1 (1)

Mà (n - 1)(n + 1) chia hết cho n - 1 (2)

Từ (1) và (2) \(\Rightarrow\)4 chia hết cho n - 1\(\Rightarrow\)n - 1 \(\in\)Ư(4) = {1 ; 2 ; 4}

\(\Rightarrow\)n \(\in\){2 ; 3 ; 5}

1)Ta có:\(2^{60}=\left(2^3\right)^{20}=8^{20}\)

\(3^{40}=\left(3^2\right)^{20}=9^{20}\)

Vì \(8^{20}< 9^{20}\Rightarrow2^{60}< 3^{40}\)

2)Gọi d là ƯCLN(n+3,2n+5)(d\(\in N\)*)

Ta có:\(n+3⋮d,2n+5⋮d\)

\(\Rightarrow2n+6⋮d,2n+5⋮d\)

\(\Rightarrow\left(2n+6\right)-\left(2n+5\right)⋮d\)

\(\Rightarrow1⋮d\)

\(\Rightarrow d=1\)

Vì ƯCLN(n+3,2n+5)=1\(\RightarrowƯC\left(n+3,2n+5\right)=\left\{1,-1\right\}\)

3)\(A=5+5^2+5^3+5^4+...+5^{98}+5^{99}\)(có 99 số hạng)

\(A=\left(5+5^2+5^3\right)+\left(5^4+5^5+5^6\right)+...+\left(5^{97}+5^{98}+5^{99}\right)\)(có 33 nhóm)

\(A=5\left(1+5+5^2\right)+5^4\left(1+5+5^2\right)+...+5^{97}\left(1+5+5^2\right)\)

\(A=5\cdot31+5^4\cdot31+...+5^{97}\cdot31\)

\(A=31\left(5+5^4+...+5^{97}\right)⋮31\left(đpcm\right)\)

6)Đặt \(A=2^1+2^2+2^3+...+2^{100}\)

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

\(2A-A=\left(2^2+2^3+2^4+...+2^{101}\right)-\left(2^1+2^2+2^3+...+2^{100}\right)\)

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

\(\Rightarrow2^1+2^2+2^3+...+2^{100}-2^{101}=2^{101}-2-2^{101}=-2\)

a)\(3n+5⋮3n-1\Rightarrow6+3n-1⋮3n-1\)

Mà \(3n-1⋮3n-1\Rightarrow6⋮3n-1\)

\(\Rightarrow3n-1\inƯ\left(6\right)\left\{-6;-3;-2;-1;1;2;3;6\right\}\)

\(\Rightarrow3n\in\left\{-5;-2;-1;0;2;3;4;7\right\}\)

\(\Rightarrow n\in\left\{\frac{-5}{3};\frac{-2}{3};\frac{-1}{3};0;\frac{2}{3};1;\frac{4}{3};\frac{7}{3}\right\}\)

Mà \(n\in N\)

\(\Rightarrow n\in\left\{0;1\right\}\)

b)\(2n+3⋮2n-1\Rightarrow4+2n-1⋮2n-1\)

Mà \(2n-1⋮2n-1\Rightarrow4⋮2n-1\)

\(\Rightarrow2n-1\in\left\{-4;-2;-1;1;2;4\right\}\)

\(\Rightarrow2n\in\left\{-3;-1;0;2;3;5\right\}\)

\(\Rightarrow n\in\left\{\frac{-3}{2};\frac{-1}{2};0;1;\frac{3}{2};\frac{5}{2}\right\}\)

Mà \(n\in N\)

\(\Rightarrow n\in\left\{0;1\right\}\)

Hok Tốt!