1/...

2/ \(=\lim\dfrac{\dfrac{1}{n\sqrt{n}}-1}{4+\dfrac{1}{n^2\sqrt{n}}}=\dfrac{0-1}{4+0}=-\dfrac{1}{4}\) (chia cả tử-mẫu cho \(n^3\))

3/ \(=\lim\dfrac{3-\left(\dfrac{1}{4}\right)^n}{2.\left(\dfrac{3}{4}\right)^n+4\left(\dfrac{1}{4}\right)^n}=\dfrac{3-0}{2.0+3.0}=\dfrac{3}{0}=+\infty\) (chia tử mẫu cho \(4^n\))

4/ \(=\lim\dfrac{2.2^n+\dfrac{4}{3}.3^n}{1-\dfrac{1}{2}.2^n+3.3^n}=\lim\dfrac{2.\left(\dfrac{2}{3}\right)^n+\dfrac{4}{3}}{\left(\dfrac{1}{3}\right)^n-\dfrac{1}{2}.\left(\dfrac{2}{3}\right)^n+3}=\dfrac{2.0+\dfrac{4}{3}}{0-\dfrac{1}{2}.0+3}=\dfrac{4}{9}\) (chia tử mẫu  cho \(3^n\))

Đề bài yêu cầu gì?

a) Theo đầu bài ta có:
\(\orbr{\begin{cases}\frac{n}{n+1}=\frac{n\left(n+4\right)}{\left(n+1\right)\left(n+4\right)}=\frac{n^2+2n+2n}{\left(n+1\right)\left(n+4\right)}\\\frac{n+1}{n+4}=\frac{\left(n+1\right)\left(n+1\right)}{\left(n+1\right)\left(n+4\right)}=\frac{n^2+2n+1}{\left(n+1\right)\left(n+4\right)}\end{cases}}\)
Nếu \(n=0\Rightarrow2n=0< 1\Rightarrow\frac{n^2+2n+2n}{\left(n+1\right)\left(n+4\right)}< \frac{n^2+2n+1}{\left(n+1\right)\left(n+4\right)}\Rightarrow\frac{n}{n+1}< \frac{n+1}{n+4}\)
Nếu \(n\ge1\Rightarrow2n\ge2>1\Rightarrow\frac{n^2+2n+2n}{\left(n+1\right)\left(n+4\right)}>\frac{n^2+2n+1}{\left(n+1\right)\left(n+4\right)}\Rightarrow\frac{n}{n+1}>\frac{n+1}{n+4}\)

\(T=\sqrt{\dfrac{2n^4-4n^3+6n^2-4n+2}{2}}+\sqrt{\dfrac{2n^4+4n^3+6n^2+4n+2}{2}}\)

\(=\sqrt{n^4-2n^3+3n^2-2n+1}+\sqrt{n^4+2n^3+3n^2+2n+1}\)

\(=\sqrt{\left(n^2-n\right)^2+2\left(n^2-n\right)+1}+\sqrt{\left(n^2+n\right)^2+2\left(n^2+n\right)+1}\)

\(=\sqrt{\left(n^2-n+1\right)^2}+\sqrt{\left(n^2+n+1\right)^2}\)

\(=n^2-n+1+n^2+n+1\)

\(=2n^2+2\ge2\)

\(T_{min}=2\) khi \(n=0\)