e: \(\Leftrightarrow2n+1\in\left\{1;-1;5;-5\right\}\)

hay \(n\in\left\{0;-1;2;-3\right\}\)

a. ĐKXĐ: \(n\ge0\)

\(lim_{n\rightarrow0}\dfrac{\sqrt{2n+1}}{\sqrt{8n}+1}=\dfrac{\sqrt{2.0+1}}{\sqrt{8.0}+1}=1\)

\(lim_{n\rightarrow+\infty}\dfrac{\sqrt{2n+1}}{\sqrt{8n}+1}=lim_{n\rightarrow+\infty}\dfrac{\sqrt{2+\dfrac{1}{n}}}{\sqrt{8}+\dfrac{1}{\sqrt{n}}}=\dfrac{1}{2}\)

b. ĐKXĐ: \(\left\{{}\begin{matrix}n\ne0\\n\le\dfrac{-1-\sqrt{21}}{2}\\n\ge\dfrac{-1+\sqrt{21}}{2}\end{matrix}\right.\)

\(lim_{n\rightarrow+\infty}\dfrac{3n+\sqrt{n^2+n-5}}{-2n}=\)\(lim_{n\rightarrow+\infty}\dfrac{3+\sqrt{1+\dfrac{1}{n}-\dfrac{5}{n^2}}}{-2}=-2\)

\(lim_{n\rightarrow-\infty}\dfrac{3n+\sqrt{n^2+n-5}}{-2n}=\)\(lim_{n\rightarrow-\infty}\dfrac{3+\sqrt{1+\dfrac{1}{n}-\dfrac{5}{n^2}}}{-2}=-1\)

a, \(lim\dfrac{\sqrt{2n+1}}{\sqrt{8n}+1}=lim\dfrac{\sqrt{n}.\sqrt{2+\dfrac{1}{n}}}{\sqrt{n}\left(\sqrt{8}+\dfrac{1}{n}\right)}=\dfrac{\sqrt{2}}{\sqrt{8}}=\dfrac{1}{2}\)

b, \(lim\dfrac{3n+\sqrt{n^2+n-5}}{-2n}\)

\(=lim\left(\dfrac{3}{2}-\dfrac{\sqrt{n^2+n-5}}{2n}\right)\)

\(=lim\left(\dfrac{3}{2}-\dfrac{n\sqrt{1+\dfrac{1}{n}-\dfrac{5}{n^2}}}{2n}\right)=\dfrac{3}{2}-\dfrac{1}{2}=1\)

\(\lim\dfrac{\sqrt{2n+1}}{\sqrt{8n}+1}=\lim\dfrac{\sqrt{n}.\sqrt{2+\dfrac{1}{n}}}{\sqrt{n}\left(\sqrt{8}+\dfrac{1}{\sqrt{n}}\right)}=\lim\dfrac{\sqrt{2+\dfrac{1}{n}}}{\sqrt{8}+\dfrac{1}{\sqrt{n}}}=\dfrac{\sqrt{2}}{\sqrt{8}}=\dfrac{1}{2}\)

\(\lim\dfrac{3n+\sqrt{n^2+n-5}}{-2n}=\lim\dfrac{n\left(3+\sqrt{1+\dfrac{1}{n}-\dfrac{5}{n^2}}\right)}{-2n}=\lim\dfrac{3+\sqrt{1+\dfrac{1}{n}-\dfrac{5}{n^2}}}{-2}=\dfrac{3+1}{-2}=-2\)

a.1

b.1

c.1

Giải thế ai hiểu nổi hả trời???

8n+1 chia hết cho 2n+1

=> 4(2n+1)-3 chia hết cho 2n+1

Vì 4(2n+1) chia hết cho 2n+1

=> 3 chia hết cho 2n+1

=> 2n+1 là ước của 3

Tự lm nốt nhá

\(8n+1⋮2n+1\)

\(\Rightarrow4\left(2n+1\right)-3⋮2n+1\)

\(\Rightarrow3⋮2n+1\)

\(\Rightarrow2n+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

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

Vậy............................

\(=\lim\left(\sqrt[]{4n^2+2n+1}-2n+2n-\sqrt[3]{8n^3-3n^2+1}\right)\)

\(=\lim\left(\dfrac{2n+1}{\sqrt[]{4n^2+2n+1}+2n}+\dfrac{3n^2-1}{4n^2+2n\sqrt[3]{8n^3-3n^2+1}+\sqrt[3]{\left(8n^3-3n^2+1\right)^2}}\right)\)

\(=\lim\left(\dfrac{2+\dfrac{1}{n}}{\sqrt[]{4+\dfrac{2}{n}+\dfrac{1}{n^2}}+2}+\dfrac{3-\dfrac{1}{n^2}}{4+2\sqrt[3]{8-\dfrac{3}{n}+\dfrac{1}{n^3}}+\sqrt[3]{\left(8-\dfrac{3}{n}+\dfrac{1}{n^3}\right)^2}}\right)\)

\(=\dfrac{2}{\sqrt[]{4}+2}+\dfrac{3}{4+2\sqrt[3]{8}+\sqrt[3]{8^2}}=...\)