2.n+5 chia hết cho n+1

=> 2n+2+3 chia hết cho n+1

=> 2(n+1)+3 chia hết cho n+1

mà 2(n+1) chia hết cho n+1

=> 3 chia hết cho n+1

=> n+1 thuộc ước của 3

=> ......................

Ta có 2n+5=2(n+1)+3

Để 2n+5 chia hết cho n+1 thì 2(n+1)+3 chia hết cho n+1

Vì 2(n+1) chia hết cho n+1 => 3 chia hết cho n+1

n thuộc N => n+1 thuộc N 

=> n+1 thuộc Ư (3)={1;3}

Nếu n+1=1 => n=0

Nếu n+1=3 => n=2

Vậy n={0;2}

\(2n+5⋮n+1\)

\(2\left(n+1\right)+3⋮n+1\)

Vì \(2\left(n+1\right)⋮n+1\)

\(3⋮n+1\)

\(\Rightarrow n+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

Ta lập bảng 

a=b(mod n) là công thức dùng để chỉ a,b có cùng số dư khi chia cho n, gọi là đồng dư thức 
Ta có các tính chất cua đồng dư thức và các tính chất sau: 
Cho x là số tự nhiên 
Nếu x lẻ thì => x^2 =1 (mod 8) 
x^2 =-1(mod 5) hoặc x^2=0(mod 5) 
Nếu x chẵn thì x^2=-1(mod 5) hoặc x^2 =1(mod 5) hoặc x^2=0(mod 5) 
Vì 2a +1 và 3a+1 là số chính phương nên ta đặt 
3a+1=m^2 
2a+1 =n^2 
=> m^2 -n^2 =a (1) 
m^2 + n^2 =5a +2 (2) 
3n^2 -2m^2=1(rút a ra từ 2 pt rồi cho = nhau) (3) 
Từ (2) ta có (m^2 + n^2 )=2(mod 5) 
Kết hợp với tính chất ở trên ta => m^2=1(mod 5); n^2=1(mod 5) 
=> m^2-n^2 =0(mod 5) hay a chia hết cho 5 
từ pt ban đầu => n lẻ =>n^2=1(mod 8) 
=> 3n^2=3(mod 8) 
=> 3n^2 -1 = 2(mod 8) 
=> (3n^2 -1)/2 =1(mod 8) 
Từ (3) => m^2 = (3n^2 -1)/2 
do đó m^2 = 1(mod 8) 
ma n^2=1(mod 8) 
=> m^2 - n^2 =0 (mod 8) 
=> a chia hết cho 8 
Ta có a chia hết cho 8 và 5 và 5,8 nguyên tố cùng nhau nên a chia hết cho 40.Vậy a là bội của 40 

Gọi 4 số chẵn cần tìm là : a ; a + 2 ; a + 4 ; a + 6

 ( a + a + 2 + a + 4  + a + 6 ) : 4 = 3

  ( 4a + 12 ) : 4 = 3

    4a + 12        = 12

    4a                = 0

      a                = 0

Vậy a = 0 ;

Tổng của 4 số đó : 3x4=12

để tổng bằng 12 thì chỉ có bốn số 0,2,4,6

đáp số : 0,2,4,6

Nguyen ngoc dat làm đúng rồi.

 VICTOR_Nobita Kun: Đề ko nói là số chẵn liên tiếp

Làm nhanh cho mình nha ( nhớ trình bày đầy đủ ) .Xin cảm ơn mọi người!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!~~~~

dãy số có n số hạng

tổng dãy số là (n + 1) x n : 2 = 465 

n x (n+1) = 930 

nhận thấy n và n + 1 là hai số tự nhiên liên tiếp 

có 30 x 31 =  930

vậy n = 30

Đó là dãy số có n số hạng

Tổng dãy số đó là (n+1)xn|2=465

nx(n+1)=930

vậy n=30

 len chua be mot canh sen

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