Tính \(n\in N\)*
\(A=\dfrac{2n^2-4n+5}{n-2}\)
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NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 1 2021
\(a=\lim\left(\dfrac{2n^3\left(5n+1\right)+\left(2n^2+3\right)\left(1-5n^2\right)}{\left(2n^2+3\right)\left(5n+1\right)}\right)\)
\(=\lim\left(\dfrac{2n^3-13n^2+3}{\left(2n^2+3\right)\left(5n+1\right)}\right)=\lim\dfrac{2-\dfrac{13}{n}+\dfrac{3}{n^3}}{\left(2+\dfrac{3}{n^2}\right)\left(5+\dfrac{1}{n}\right)}=\dfrac{2}{2.5}=\dfrac{1}{5}\)
\(b=\lim\left(\dfrac{n-2}{\sqrt{n^2+n}+\sqrt{n^2+2}}\right)=\lim\dfrac{1-\dfrac{2}{n}}{\sqrt{1+\dfrac{1}{n}}+\sqrt{1+\dfrac{2}{n}}}=\dfrac{1}{2}\)
\(c=\lim\dfrac{\sqrt{1+\dfrac{3}{n^3}-\dfrac{2}{n^4}}}{2-\dfrac{2}{n}+\dfrac{3}{n^2}}=\dfrac{1}{2}\)
\(d=\lim\dfrac{\sqrt{1-\dfrac{4}{n}}-\sqrt{4+\dfrac{1}{n^2}}}{\sqrt{3+\dfrac{1}{n^2}}-1}=\dfrac{1-2}{\sqrt{3}-1}=-\dfrac{1+\sqrt{3}}{2}\)
24 tháng 11 2023
1: \(\lim\limits_{n\rightarrow\infty}\dfrac{6n-8}{n-1}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{6-\dfrac{8}{n}}{1-\dfrac{1}{n}}=\dfrac{6-0}{1-0}\)
\(=\dfrac{6}{1}=6\)
2: \(\lim\limits_{n\rightarrow\infty}\dfrac{n^2+5n-3}{4n^3-2n+5}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(1+\dfrac{5}{n}-\dfrac{3}{n^2}\right)}{n^3\left(4-\dfrac{2}{n^2}+\dfrac{5}{n^3}\right)}\)
\(=\lim\limits_{n\rightarrow\infty}\left(\dfrac{1}{n}\cdot\dfrac{1+\dfrac{5}{n}-\dfrac{3}{n^2}}{\left(4-\dfrac{2}{n^2}+\dfrac{5}{n^3}\right)}\right)\)
=0
MH
13 tháng 10 2023
1) \(\lim\limits_{n\rightarrow\infty}\dfrac{6n-8}{n-1}=\lim\limits_{n\rightarrow\infty}\dfrac{2n\left(1-\dfrac{4}{n}\right)}{n\left(1-\dfrac{1}{n}\right)}=2\)
2) \(\lim\limits_{n\rightarrow\infty}\dfrac{n^2+5n-3}{4n^3-2n+5}=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(1+\dfrac{5}{n}-\dfrac{3}{n^2}\right)}{n^3\left(4-\dfrac{2}{n^2}+\dfrac{5}{n^3}\right)}=\dfrac{1}{4n}=\infty\)
3) \(\lim\limits_{n\rightarrow\infty}\left(-2n^5+4n^4-3n^2+4\right)=\lim\limits_{n\rightarrow\infty}n^5\left(-2+\dfrac{4}{n}-\dfrac{3}{n^2}+\dfrac{4}{n^5}\right)=-2n^5=-\infty\)
TN
29 tháng 4 2017
BÀi 1
Để A \(\in\) Z
=>\(\left(n+2\right)⋮\left(n-5\right)\)
=>\([\left(n-5\right)+7]⋮\left(n-5\right)\)
=>\(7⋮\left(n-5\right)\)
=>\(n-5\in\left\{1;7;-1;-7\right\}\)
=>\(n\in\left\{6;13;4;-2\right\}\)
Vậy \(n\in\left\{6;13;4;-2\right\}\)
11 tháng 5 2022
a, \(A=\dfrac{5n-4-4n+5}{n-3}=\dfrac{n+1}{n-3}=\dfrac{n-3+4}{n-3}=1+\dfrac{4}{n-3}\Rightarrow n-3\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
| n-3 | 1 | -1 | 2 | -2 | 4 | -4 |
| n | 4 | 2 | 5 | 1 | 7 | -1 |
11 tháng 5 2022
a.\(A=\dfrac{2n+1}{n-3}+\dfrac{3n-5}{n-3}-\dfrac{4n-5}{n-3}\)
\(A=\dfrac{2n+1+3n-5-4n+5}{n-3}\)
\(A=\dfrac{n+1}{n-3}\)
\(A=\dfrac{n-3}{n-3}+\dfrac{4}{n-3}\)
\(A=1+\dfrac{4}{n-3}\)
Để A nguyên thì \(\dfrac{4}{n-3}\in Z\) hay \(n-3\in U\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
n-3=1 --> n=4
n-3=-1 --> n=2
n-3=2 --> n=5
n-3=-2 --> n=1
n-3=4 --> n=7
n-3=-4 --> n=-1
Vậy \(n=\left\{4;2;5;7;1;-1\right\}\) thì A nhận giá trị nguyên
b.hemm bt lèm:vv
HT
17 tháng 1 2021
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
\(=lim\dfrac{8n^6}{3n^6}=\dfrac{8}{3}\)
20 tháng 11 2023
\(\lim\limits\dfrac{\sqrt{4n^2+1}+2n-1}{\sqrt{n^2+4n+1}+n}\)
\(=\lim\limits\dfrac{\sqrt{4+\dfrac{1}{n^2}}+2-\dfrac{1}{n}}{\sqrt{1+\dfrac{4}{n}+\dfrac{1}{n^2}}+1}=\dfrac{2+2}{1+1}=\dfrac{4}{2}=2\)
\(\lim\limits\left[\sqrt{n}\left(\sqrt{n+1}-n\right)\right]\)
\(=\lim\limits\left[\sqrt{n^2+n}-\sqrt{n^3}\right]\)
\(=\lim\limits\dfrac{n^2+n-n^3}{\sqrt{n^2+n}+\sqrt{n^3}}\)
\(=\lim\limits\dfrac{n^3\left(-1+\dfrac{1}{n}+\dfrac{1}{n^2}\right)}{\sqrt{n^3\left(\dfrac{1}{n}+\dfrac{1}{n^2}\right)}+\sqrt{n^3}}\)
\(=\lim\limits\dfrac{n^3\left(-1+\dfrac{1}{n}+\dfrac{1}{n^2}\right)}{\sqrt{n^3}\left(\sqrt{\dfrac{1}{n}+\dfrac{1}{n^2}}+1\right)}\)
\(=\lim\limits\dfrac{n\sqrt{n}\left(-1+\dfrac{1}{n}+\dfrac{1}{n^2}\right)}{\sqrt{\dfrac{1}{n}+\dfrac{1}{n^2}}+1}\)
\(=-\infty\) vì \(\left\{{}\begin{matrix}lim\left(n\sqrt{n}\right)=+\infty\\lim\left(\dfrac{-1+\dfrac{1}{n}+\dfrac{1}{n^2}}{\sqrt{\dfrac{1}{n}+\dfrac{1}{n^2}}+1}\right)=-\dfrac{1}{1}=-1< 0\end{matrix}\right.\)
\(A=\dfrac{2n^2-4n+5}{n-2}=\dfrac{2n\left(n-2\right)+5}{n-2}=\dfrac{2n\left(n-2\right)}{n-2}+\dfrac{5}{n-2}=2+\dfrac{5}{n-2}\)
Ta có: \(2\) là số nguyên nên để \(A\) nguyên
Suy ra \(5⋮n-2\Rightarrow n-2\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)
\(\Rightarrow n\in\left\{3;1;7\right\}\left(n\in N\text{*}\right)\)