\(a=\lim4^n\left(1-\left(\dfrac{3}{4}\right)^n\right)=+\infty.1=+\infty\)

\(b=\lim\left(4^n+2.2^n+1-4^n\right)=\lim2^n\left(2+\dfrac{1}{2^n}\right)=+\infty.2=+\infty\)

\(c=limn^3\left(\sqrt{\dfrac{2}{n}-\dfrac{3}{n^4}+\dfrac{11}{n^6}}-1\right)=+\infty.\left(-1\right)=-\infty\)

\(d=\lim n\left(\sqrt{2+\dfrac{1}{n^2}}-\sqrt{3-\dfrac{1}{n^2}}\right)=+\infty\left(\sqrt{2}-\sqrt{3}\right)=-\infty\)

\(e=\lim\dfrac{3n\sqrt{n}+1}{\sqrt{n^2+3n\sqrt{n}+1}+n}=\lim\dfrac{3\sqrt{n}+\dfrac{1}{n}}{\sqrt{1+\dfrac{3}{\sqrt{n}}+\dfrac{1}{n^2}}+1}=\dfrac{+\infty}{2}=+\infty\)

a: \(\left(n^2+3n-1\right)\left(n+2\right)-n^3+2\)

\(=n^3+2n^2+3n^2+6n-n-2+n^3+2\)

\(=5n^2+5n=5\left(n^2+n\right)⋮5\)

b: \(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)\)

\(=6n^2+30n+n+5-6n^2+3n-10n+5\)

\(=24n+10⋮2\)

d: \(=\left(n+1\right)\left(n^2+2n\right)\)

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

\(------huongdan-----\)

\(Taco:\)

\(\left(3n-2n\right)⋮n+1\Leftrightarrow n⋮n+1\Leftrightarrow\left(n+1\right)-n⋮n+1\Leftrightarrow1⋮n+1\)

\(\Leftrightarrow n+1\in\left\{-1;1\right\}\Leftrightarrow n\in\left\{-2;0\right\}\)

\(b,2n-4⋮n+2\Leftrightarrow2n+4-2n+4⋮2n+4\Leftrightarrow8⋮2n+4\)

dễ thấy: 2n+4 chẵn => 2n+4 là ước chẵn của 8

\(\Rightarrow2n+4\in\left\{2;4;8;-2;-4;-8\right\}\Rightarrow2n\in\left\{-2;0;4;-6;-8;-12\right\}\)

\(\Rightarrow n\in\left\{-1;0;2;-3;-4;-6\right\}\)

\(2n-4⋮n+2\)

\(\Rightarrow2n+4-8⋮n+2\)

\(\Rightarrow2\left(n+2\right)+8⋮n+2\)

\(\Rightarrow n+2\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

bn tụ lập bảng ha ~ 

a)\(3n-2n⋮n+1\)

\(\Rightarrow n⋮n+1\)

\(\Rightarrow n+1-1⋮n+1\Rightarrow1⋮n+1\orbr{\begin{cases}n+1=1\\n+1=-1\end{cases}}\Rightarrow\orbr{\begin{cases}n=0\\n=-2\end{cases}}\)

b)\(2n-4⋮n+2\Rightarrow2n+4-8⋮n+2\)

\(\Rightarrow2\left(n+2\right)-8⋮n+2\Rightarrow8⋮n+2\)

\(+,n+2=1\Rightarrow n=-2\)

\(+,n+2=-1\Rightarrow n=-3\)

\(+,n+2=2\Rightarrow n=0\)

\(+,n+2=-2\Rightarrow n=-4\)

\(+,n+2=4\Rightarrow n=2\)

\(+,n+2=-4\Rightarrow n=-6\)

Dang này thì cứ chọn số hạng có mũ cao nhất trên tử và mẫu là được. Nó là ngắt vô cùng lớn hay bé gì đấy

\(=lim\dfrac{8n^6}{3n^6}=\dfrac{8}{3}\)

a/ \(=lim\frac{\left(-\frac{2}{3}\right)^n+1}{-2.\left(-\frac{2}{3}\right)^n+3}=\frac{1}{3}\)

b/ \(=lim\frac{\left(2-\frac{1}{n}\right)\left(1+\frac{1}{n}\right)\left(3+\frac{4}{n}\right)}{\left(\frac{5}{n}-6\right)^3}=\frac{2.1.3}{\left(-6\right)^3}=-\frac{1}{36}\)

c/ \(=lim\frac{5n+3}{\sqrt{n^2+5n+1}+\sqrt{n^2-2}}=\frac{5+\frac{3}{n}}{\sqrt{1+\frac{5}{n}+\frac{1}{n^2}}+\sqrt{1-\frac{2}{n}}}=\frac{5}{1+1}=\frac{5}{2}\)

d/ \(=lim\frac{5.\left(\frac{1}{2}\right)^n-6}{4.\left(\frac{1}{3}\right)^n+1}=\frac{-6}{1}=-6\)

e/ \(=-n^3\left(2+\frac{3}{n}-\frac{5}{n^2}+\frac{2020}{n^3}\right)=-\infty.2=-\infty\)

a) Vì 3\(⋮\)n

=> n\(\in\)Ư₍₃₎={ 1; 3 }

Vậy, n=1 hoặc n=3

A:    n=3;1                  E:     n=2

B:     n=6;2                  F:    n=2

c:     n=1                     G:     n=2

D:    n=2                      H:     n=5