\(n^2-2n-22\) là bội \(n+3\)

\(\Rightarrow n^2-2n-22⋮n+3\)

\(\Rightarrow n^2+3n-5n-22⋮n+3\)

\(\Rightarrow n\left(n+3\right)-5n-22⋮n+3\)

Ta có: \(n\left(n+3\right)⋮n+3\) nên để \(n^2-2n-22⋮n+3\)

thì \(-5n-22⋮n-3\)\(\Rightarrow-5\left(n-3\right)-7⋮n-3\)

Mà \(-5\left(n-3\right)⋮n-3\) suy ra \(-7⋮n-3\)

\(\Rightarrow n-3\inƯ\left(-7\right)=\left\{1;-1;7;-7\right\}\)

\(\Rightarrow n\in\left\{4;2;10;-4\right\}\)

n2−2n−22 là bội $n+3$

⇒�2−2�−22⋮�+3⇒n2−2n−22⋮n+3

⇒�2+3�−5�−22⋮�+3⇒n2+3n−5n−22⋮n+3

$\Rightarrow n\left(n+3\right)-5n-22⋮n+3$

Ta có: $n\left(n+3\right)⋮n+3$ nên để �2−2�−22⋮�+3n2−2n−22⋮n+3

thì $-5n-22⋮n-3$$\Rightarrow -5\left(n-3\right)-7⋮n-3$

Mà $-5\left(n-3\right)⋮n-3$ suy ra $-7⋮n-3$

$\Rightarrow n-3\in Ư\left(-7\right)=\left\{1;-1;7;-7\right\}$

$\Rightarrow n\in \left\{4;2;10;-4\right\}$

viết lại đề đi, mik ko hiểu

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Đề bài yêu cầu gì vậy bạn?

Bài 1 :

A = 1² + 2² + 3² +....+n² 

A = 1² + 2.(1+1) + 3.(2 +1) + 4.( 3 +1) +.....+n(n-1 + 1)

A = 1 + 1.2 + 2 + 2.3 + 3 + 3.4 + 4 +.....+ n.(n-1) + n

A = ( 1 + 2 + 3 + 4 +....+n) + ( 1.2 + 2.3 + 3.4 +....+(n-1).n

A = (n+1).{(n-1):n+1)/2 +1/3.[1.2.3 +2.3.3 +.....+(n-1)n.3]

A = (n+1).n/2+1/3.[1.2.3 +2.3.(4-1)+ ...+(n-1).n [(n+1) - (n -2)]

A = (n+1)n/2+1/3.( 1.2.3 + 2.3.4 -1.2.3 +..+ (n-1)n(n+1)- (n-2)(n-1)n)

A =(n+1)n/2 + 1/3.(n-1)n(n+1)

A = n(n+1)[1/2 + 1/3 .(n-1)]

A = n.(n+1) \(\dfrac{3+2n-2}{6}\)

A= n.(n+1)(2n+1)/6

Bài 2 : 

a, (x+1) +(x+2) + (x+3)+...+(x+10) = 5070

    (x+10 +x+1).{( x+10 - x -1): 1 +1):2  = 5070

    (2x + 11)10 : 2 = 5070 

     ( 2x + 11)5 = 5070

      2x+ 11 = 5070:5

         2x = 1014 - 11

        2x =   1003

          x = 1003 :2

          x = 501,5 

        b, 1 + 2 + 3 +...+x = 820

           ( x + 1)[ (x-1):1 +1] : 2 = 820

           (x +1).x = 820 x 2

           (x +1).x = 1640

            (x +1) .x = 40 x 41

                 x = 40 

Câu hỏi của Ngọn Gió Thần Sầu - Toán lớp 6 - Học toán với OnlineMath

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9: \(\Leftrightarrow n^2+n+3n+2+1⋮n+1\)

\(\Leftrightarrow n+1\in\left\{1;-1\right\}\)

hay \(n\in\left\{0;-2\right\}\)

10: \(\Leftrightarrow n^2+4n+4-2⋮n+2\)

\(\Leftrightarrow n+2\in\left\{1;-1;2;-2\right\}\)

hay \(n\in\left\{-1;-3;0;-4\right\}\)

11: \(\Leftrightarrow n^2-2n+1+2⋮n-1\)

\(\Leftrightarrow n-1\in\left\{1;-1;2;-2\right\}\)

hay \(n\in\left\{2;0;3;-1\right\}\)

9) n² + 3n +3 ⋮ n +1

10) n² + 4n + 2 ⋮ n +2

11)n² - 2n + 3 ⋮ n - 1

a: Ta có: \(3n+2⋮n-1\)

\(\Leftrightarrow n-1\in\left\{1;-1;5;-5\right\}\)

hay \(n\in\left\{2;0;6;-4\right\}\)

1. Áp dụng công thức tổng cấp số nhân:

\(S_n=u_1.\dfrac{q^n-1}{q-1}=2.\dfrac{2^n-1}{2-1}=2.\left(2^n-1\right)=2^{n+1}-2\)

2. \(\left\{{}\begin{matrix}u_2+u_5=12\\u_4+u_8=22\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(u_1+d\right)+\left(u_1+4d\right)=12\\\left(u_1+3d\right)+\left(u_1+7d\right)=22\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2u_1+5d=12\\2u_1+10d=22\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u_1=1\\d=2\end{matrix}\right.\)

\(\Rightarrow u_n=u_1+\left(n-1\right)d=1+\left(n-1\right)2=2n-1\)

\(\Rightarrow S_n=\dfrac{n\left(u_1+u_n\right)}{2}=\dfrac{n\left(1+2n-1\right)}{2}=n^2\)

3. \(\left\{{}\begin{matrix}u_1+u_2=4\\u_4+u_1=28\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u_1+u_1q=4\\u_1q^3+u_1=28\end{matrix}\right.\)

\(\Rightarrow\dfrac{q^3+1}{q+1}=\dfrac{28}{4}\Rightarrow q^2-q+1=7\)

\(\Rightarrow q^2-q-6=0\Rightarrow\left[{}\begin{matrix}q=3\\q=-2\end{matrix}\right.\)