a, Ta có 8n - 59 = ( 2n -16 ) + ( 2n -16 ) + ( 2n - 16 ) + ( 2n - 16 ) + 5

2n - 16 luôn luôn chia hết cho 2n - 16 

=> 4.(2n-16) chia hết cho 2n-16 <=> 5 chia hết cho 2n - 16

=> 2n - 16 thuộc Ư(5) = { 1;-1;5;-5 }

Tự làm nốt

b, tương tự 

c, 6n - 46 = (2n-18) + (2n-18) + (2n-18) + 8

... Tiếp tục :))

a ,\(8n-59⋮2n-16\)

Mà \(2n-16⋮2n-16\) 

\(\Rightarrow4\left(2n-16\right)⋮2n-16\)

\(\Rightarrow8n-64⋮2n-16\) 

\(\Rightarrow\left(8n-59\right)-\left(8n-64\right)⋮2n-16\) 

\(\Rightarrow8n-59-8n+64⋮2n-16\) 

\(\Rightarrow5⋮2n-16\) 

\(\Rightarrow2n-16\inƯ\left(5\right)\) 

\(\Rightarrow2n-16\in\left\{\pm1;\pm5\right\}\) 

\(\Rightarrow2n\in\left\{17;15;21;11\right\}\) 

\(\Rightarrow\) KHÔNG CÓ SỐ NÀO THỎA MÃN CỦA 2n 

\(\Rightarrow x\in\varnothing\)

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

\(=lim\left(n\left(1-2\sqrt{n}\sqrt{2+\frac{2}{n}+\frac{1}{2n^2}}\right)\right)=\infty\times\left(-\infty\right)=-\infty\)

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

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

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

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

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

\(=\frac{-8+0}{4+2\sqrt[3]{8+8+0}+\sqrt[3]{\left(8+0+0\right)^2}}=\frac{-2}{3}\)

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

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

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

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

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

- Với \(n=0\) không thỏa mãn

- Với \(n=1\) không thỏa mãn

- Với \(n=2\Rightarrow2^n+8n+5=25\) là số chính phương (thỏa mãn)

- Với \(n>2\Rightarrow2^n⋮8\Rightarrow2^n+8n+5\) chia 8 dư 5

Mà 1 SCP chia 8 chỉ có các số dư là 0, 1, 4 nên \(2^n+8n+5\) ko thể là SCP 

Vậy \(n=2\) là giá trị duy nhất thỏa mãn yêu cầu

Để 8n - 9 chia hết cho 2n + 5

=> ( 8n + 20 ) - 29 chia hết cho 2n + 5

=> 4(2n + 5) - 29 chia hết cho 2n + 5

=> 29 chia hết cho 2n + 5

=> 2n + 5 thuộc Ư(29) = { - 29 ; - 1 ; 1 ; 29 }

Vậy n thuộc {  - 19 ; -3 ; -2 ; 12 }