Để (2^n-1);7 thì nó phải thuộc U(7) =1:-1;7;-7

Vậy n=3 thì   (2^n-1);7

có ai giúp mik với

a, \(n^2+2n-4=n^2+2n-15+11=\left(n-3\right)\left(n-5\right)+11\)

Để \(n^2+2n-4⋮11\Leftrightarrow\left(n-3\right)\left(n+5\right)⋮11\Leftrightarrow\left[{}\begin{matrix}n-3⋮11\\n+5⋮11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}n=BS11+3\\n=BS11-5\end{matrix}\right.\)

c,\(\dfrac{n^3-n^2+2n+7}{n^2+1}=\dfrac{n^3+n-n^2-1+n+8}{n^2+1}=\dfrac{n\left(n^2+1\right)-\left(n^2+1\right)+n+8}{n^2+1}=n-1+\dfrac{n+8}{n^2+1}\)

Để \(n^3-n^2+2n+7⋮n^2+1\Leftrightarrow n+8⋮n^2+1\)

\(\Rightarrow\left(n+8\right)\left(n-8\right)⋮n^2+1\Rightarrow n^2-64⋮n^2+1\)

\(\Rightarrow n^2+1-65⋮n^2+1\Rightarrow65⋮n^2+1\)

\(\Rightarrow n^2+1\inƯ\left(65\right)=\left\{\pm1;\pm5;\pm13;\pm65\right\}\)

Mà \(n^2+1\ge1\Rightarrow n^2+1\in\left\{1;5;13;65\right\}\)

\(\Rightarrow n\in\left\{0;\pm2;\sqrt{12};\pm8\right\}\)

Câu c ý tưởng thì hay đó, mỗi tội thiếu bước thử lại

a)

\(55^{n+1}-55^n\\ =55^n.55-55^n\\ =55^n\left(55-1\right)\\ =55^n.54⋮54\\ \RightarrowĐpcm\)

b)

\(n^2\left(n+1\right)+2n\left(n+1\right)\\ =\left(n+1\right)\left(n^2+2n\right)\\ =n\left(n+1\right)\left(n+2\right)⋮6\\ \)

c)

\(2^{n+2}+2^{n+1}+2^n\\ =2^n.2^2+2^n.2+2^n\\ =2^n\left(4+2+1\right)\\ =2^n.7⋮7\)

ko bít

ko biết nói làm j