Câu 4.

\(\lim \left( {{n^2}\sin \dfrac{{n\pi }}{5} - 2{n^3}} \right) = \lim {n^3}\left( {\dfrac{{\sin \dfrac{{n\pi }}{5}}}{n} - 2} \right) = - \infty \)

Vì \(\lim {n^3} = + \infty ;\lim \left( {\dfrac{{\sin \dfrac{{n\pi }}{5}}}{n} - 2} \right) = - 2 \)

\(\left| {\dfrac{{\sin \dfrac{{n\pi }}{5}}}{n}} \right| \le \dfrac{1}{n};\lim \dfrac{1}{n} = 0 \Rightarrow \lim \left( {\dfrac{{\sin \dfrac{{n\pi }}{5}}}{n} - 2} \right) = - 2\)

Câu 5.

Ta có: \(\left\{ \begin{array}{l} 0 \le \left| {{u_n}} \right| \le \dfrac{1}{{{n^2} + 1}} \le \dfrac{1}{n} \to 0\\ 0 \le \left| {{v_n}} \right| \le \dfrac{1}{{{n^2} + 2}} \le \dfrac{1}{n} \to 0 \end{array} \right. \to \lim {u_n} = \lim {v_n} = 0 \to \lim \left( {{u_n} + {v_n}} \right) = 0\)

a) \(\dfrac{81}{\left(-3\right)^n}=-243\)

\(\dfrac{\left(-3\right)^4}{\left(-3\right)^n}=\left(-3\right)^5\)

\(\left(-3\right)^n=\dfrac{\left(-3\right)^4}{\left(-3\right)^5}=\left(-3\right)^{-1}\)

n = -1

Vậy n = -1

b) \(\dfrac{25}{5^n}=5\)

\(\dfrac{5^2}{5^n}=5^1\)

\(5^n=\dfrac{5^2}{5^1}=5^1\)

n = 1

Vậy n = 1

c) \(\dfrac{1}{2}\cdot2^n+4\cdot2^n=9\cdot2^5\)

\(2^{n-1}+4\cdot2^{n-1}\cdot2=9\cdot2^5\)

\(2^{n-1}+8\cdot2^{n-1}=9\cdot2^5\)

\(\left(8+1\right)\cdot2^{n-1}=9\cdot2^5\)

\(9\cdot2^{n-1}=9\cdot2^5\)

\(2^{n-1}=2^5\cdot\dfrac{9}{9}=2^5\)

n - 1 = 5

n = 5 + 1 = 6

Vậy n = 6

a) 81/(-3)ⁿ = -243

(-3)ⁿ = 81 : (-243)

(-3)ⁿ = -1/3

n = -1

b) 25/5ⁿ = 5

5ⁿ = 25 : 5

5ⁿ = 5

n = 1

c) 1/2 . 2ⁿ + 4 . 2ⁿ = 9 . 2⁵

2ⁿ . (1/2 + 4) = 9 . 32

2ⁿ . 9/2 = 288

2ⁿ = 288 : 9/2

2ⁿ = 64

2ⁿ = 2⁶

n = 6

Bài 2: 

1: \(5^n+5^{n+2}=650\)

\(\Leftrightarrow5^n\cdot26=650\)

\(\Leftrightarrow5^n=25\)

hay x=2

2: \(32^{-n}\cdot16^n=1024\)

\(\Leftrightarrow\dfrac{1}{32^n}\cdot16^n=1024\)

\(\Leftrightarrow\left(\dfrac{1}{2}\right)^n=1024\)

hay n=-10

13: \(9\cdot27^n=3^5\)

\(\Leftrightarrow3^{3n}=3^5:3^2=3^3\)

=>3n=3

hay n=1

a) \(\lim \frac{{2{n^2} + 6n + 1}}{{8{n^2} + 5}} = \lim \frac{{{n^2}\left( {2 + \frac{6}{n} + \frac{1}{{{n^2}}}} \right)}}{{{n^2}\left( {8 + \frac{5}{{{n^2}}}} \right)}} = \lim \frac{{2 + \frac{6}{n} + \frac{1}{n}}}{{8 + \frac{5}{n}}} = \frac{2}{8} = \frac{1}{4}\)

b) \(\lim \frac{{4{n^2} - 3n + 1}}{{ - 3{n^3} + 6{n^2} - 2}} = \lim \frac{{{n^3}\left( {\frac{4}{n} - \frac{3}{{{n^2}}} + \frac{1}{{{n^3}}}} \right)}}{{{n^3}\left( { - 3 + \frac{6}{n} - \frac{2}{{{n^3}}}} \right)}} = \lim \frac{{\frac{4}{n} - \frac{3}{{{n^2}}} + \frac{1}{{{n^3}}}}}{{ - 3 + \frac{6}{n} - \frac{2}{{{n^3}}}}} = \frac{{0 - 0 + 0}}{{ - 3 + 0 - 0}} = 0\).

c) \(\lim \frac{{\sqrt {4{n^2} - n + 3} }}{{8n - 5}} = \lim \frac{{n\sqrt {4 - \frac{1}{n} + \frac{3}{{{n^2}}}} }}{{n\left( {8 - \frac{5}{n}} \right)}} = \frac{{\sqrt {4 - 0 + 0} }}{{8 - 0}} = \frac{2}{8} = \frac{1}{4}\).

d) \(\lim \left( {4 - \frac{{{2^{{\rm{n}} + 1}}}}{{{3^{\rm{n}}}}}} \right) = \lim \left( {4 - 2 \cdot {{\left( {\frac{2}{3}} \right)}^{\rm{n}}}} \right) = 4 - 2.0 = 4\).

e) \(\lim \frac{{{{4.5}^{\rm{n}}} + {2^{{\rm{n}} + 2}}}}{{{{6.5}^{\rm{n}}}}} = \lim \frac{{{{4.5}^{\rm{n}}} + {2^2}{{.2}^{\rm{n}}}}}{{{{6.5}^{\rm{n}}}}} = \lim \frac{{{5^n}.\left[ {4 + 4.{{\left( {\frac{2}{5}} \right)}^{\rm{n}}}} \right]}}{{{{6.5}^n}}} = \lim \frac{{4 + 4.{{\left( {\frac{2}{5}} \right)}^{\rm{n}}}}}{6} = \frac{{4 + 4.0}}{6} = \frac{2}{3}\).

g) \(\lim \frac{{2 + \frac{4}{{{n^3}}}}}{{{6^{\rm{n}}}}} = \lim \left( {2 + \frac{4}{{{{\rm{n}}^3}}}} \right).\lim {\left( {\frac{1}{6}} \right)^{\rm{n}}} = \left( {2 + 0} \right).0 = 0\).

đúng là ko có bài nào dễ trong ngày hôm nay

Bạn ghi nhỏ lại nhé. Hơn nũa bạn nên tách riêng từng câu hỏi, làm vầy nhiều lắm

a) Ta co :1/5^2+1/6^2+1/7^2+...+1/100^2<1/4.5+1/5.6+1/6.7+...+1/99.100

Dat A=1/4.5+1/5.6+...+1/99.100.   B=1/5^2+1/6^2+...+1/100^2

A=1/4-1/5+1/5-1/6+1/6-1/7+...+1/99-1/100

=1/4-1/100=6/25

Ma1/6<6/25<1/4.Ta lại cóA<6/25    Vậy:1/6<1/5^2+1/6^2+1/7^2+...+1/100^2<1/4

$n$ tiến đến đâu vậy bạn?

Câu 2:

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n(n+1)}=\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{(n+1)-n}{n(n+1)}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...\frac{1}{n}-\frac{1}{n+1}\)

\(=1-\frac{1}{n+1}\)

\(\Rightarrow \lim_{n\to \infty}(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n(n+1)})=\lim_{n\to \infty}(1-\frac{1}{n+1})=1-\lim_{n\to \infty}\frac{1}{n+1}=1-0=1\)

Câu 3:

Ta biết rằng $\lim_{x\to \infty}\frac{1}{x}=0\Rightarrow \lim_{x\to \infty}\frac{a}{x}=0$ với $a\in\mathbb{R}$

Do đó:

$\lim_{n\to \infty}\frac{1}{n^2}=0$

$\lim_{n\to \infty}\frac{2}{n^2}=0$

.....

$\lim_{n\to \infty}\frac{2n-1}{n^2}=\lim_{n\to \infty}(\frac{2}{n}-\frac{1}{n^2})=\lim_{n\to \infty}\frac{2}{n}-\lim_{n\to \infty}\frac{1}{n^2}=0-0=0$

Do đó:

$\lim_{n\to \infty}(\frac{1}{n^2}+...+\frac{2n-1}{n^2})=\lim_{n\to \infty}\frac{1}{n^2}+....+\lim_{n\to \infty}\frac{2n-1}{n^2}=0+0+...+0=0$