2ⁿ⁺³ + 2ⁿ = 144

<=> 2ⁿ(2³ + 1 ) = 144

<=> 2ⁿ.9 = 144

<=> 2ⁿ = 16

<=> 2ⁿ = 2⁴

<=> n= 4

         2ⁿ+ 4.2 ⁿ⁺¹ = 9.4³

<=> 2ⁿ + 2ⁿ⁺³ = 9.2⁶ 

<=> 2ⁿ( 1 + 2³ ) = 9.2⁶

<=> 2ⁿ.9 = 9.2⁶

<=> 2ⁿ = 2⁶

<=> n = 6

     #_W  

Tìm n

a) 2^{n + 3} + 2ⁿ = 144

=> 2ⁿ.2³ + 2ⁿ = 144

=> 2ⁿ.8 + 2ⁿ = 144

=> 2ⁿ.(8 + 1) = 144

=> 2ⁿ.9 = 144

=> 2ⁿ = 144 : 9

=> 2ⁿ = 16

=> 2ⁿ = 2⁴

=> n = 4

Vậy n = 4

b) 2ⁿ + 4.2^{n + 1} = 9.4³

=> 2ⁿ + 2².2^{n + 1} = 9.(2²)³

=> 2ⁿ + 2^{n + 3} = 9.2^2.3

=> 2ⁿ + 2ⁿ.2³ = 9.2⁶

=> 2ⁿ + 2ⁿ.8 = 9.2⁶

=> 2ⁿ.(1 + 8) = 9.2⁶

=> 2ⁿ.9 = 9.2⁶

=> 2ⁿ = 2⁶

=> n = 6

Vậy n = 6

a) 2n + 3 + 2n = 144

=> 2n.23 + 2n = 144

=> 2n.8 + 2n = 144

=> 2n.(8 + 1) = 144

=> 2n.9 = 144

=> 2n = 144 : 9

=> 2n = 16

=> 2n = 24

=> n = 4

Vậy n = 4

b) 2n + 4.2n + 1 = 9.43

=> 2n + 22.2n + 1 = 9.(22)3

=> 2n + 2n + 3 = 9.22.3

=> 2n + 2n.23 = 9.26

=> 2n + 2n.8 = 9.26

=> 2n.(1 + 8) = 9.26

=> 2n.9 = 9.26

=> 2n = 26

=> n = 6

Vậy n = 6

1.

\(\lim (n^3+4n^2-1)=\infty\) khi $n\to \infty$

2. 

\(\lim\limits_{n\to -\infty} \frac{(n+1)\sqrt{n^2-n+1}}{3n^2+n}=\lim\limits_{n\to -\infty}\frac{-\frac{n+1}{n}.\sqrt{\frac{n^2-n+1}{n^2}}}{3+\frac{1}{n}}\\ =\lim\limits_{n\to -\infty}\frac{-(1+\frac{1}{n})\sqrt{1-\frac{1}{n}+\frac{1}{n^2}}}{3+\frac{1}{n}}=\frac{-1}{3}\)

\(\lim\limits_{n\to +\infty} \frac{(n+1)\sqrt{n^2-n+1}}{3n^2+n}=\lim\limits_{n\to +\infty}\frac{\frac{n+1}{n}.\sqrt{\frac{n^2-n+1}{n^2}}}{3+\frac{1}{n}}\\ =\lim\limits_{n\to +\infty}\frac{(1+\frac{1}{n})\sqrt{1-\frac{1}{n}+\frac{1}{n^2}}}{3+\frac{1}{n}}=\frac{1}{3}\)

3.

\(\lim \frac{1+2+...+n}{2n^2}=\lim \frac{n(n+1)}{4n^2}=\lim \frac{n^2+n}{4n^2}\\ =\lim (\frac{1}{4}+\frac{1}{4n})=\frac{1}{4}\)

4.

\(\lim \frac{3^n-4.2^{n-1}-10}{7.2^n+4^n}=\lim \frac{(\frac{3}{4})^n-(\frac{2}{4})^{n-1}-\frac{10}{4^n}}{7(\frac{2}{4})^n+1}\\ =\lim \frac{(\frac{3}{4})^n-(\frac{1}{2})^{n-1}-\frac{10}{4^n}}{7(\frac{1}{2})^n+1}\\ =\frac{0-0-0}{7.0+1}=0\)

a, 

5ⁿ + 5^{n + 2} = 650

=> 5ⁿ + 5ⁿ.5² = 650

=> 5ⁿ(1 + 5²) = 650

=> 5ⁿ.26 = 650

=> 5ⁿ = 25

=> n = 2

a) 5ⁿ +5ⁿ⁺² = 650

5ⁿ + 5ⁿ.5² = 650

5ⁿ.(1+25 ) = 650

5ⁿ.26= 650

5ⁿ = 25 = 5² 

=> n = 2

b) 3ⁿ⁺³ +5.3ⁿ = 864

3ⁿ.3³ +5.3ⁿ = 864

3ⁿ.(3³+5) = 864

3ⁿ.32 = 864

3ⁿ = 27 = 3³

=> n = 3

các bài cn lại bn dựa vào mak lm nha!
 

ggggggggggggggggggggggggg

a) 5ⁿ + 5ⁿ⁺² = 650   

=> 5ⁿ+5ⁿ⁺²=5⁴ +5²

=> n+n+2 = 4+2

=>2n +2 = 6

=> n=2

  b) 3ⁿ + 5.3ⁿ= 864 

=> 3ⁿ .(1+5) = 864

=> 3ⁿ = 864 :6

=> 3ⁿ =144

=> 3ⁿ =3²+3³+3⁴-3

=> n=2+3+4-3=6

 c ) 5ⁿ⁺³ - 5ⁿ⁺¹= (125)⁴ . 120

 => 5ⁿ⁺³ - 5^{n+ =} 5¹². ( 5^3 -5)

=> n+3 -n = 12.2

=> 3=14 ( vô lí )

=> không tồn tại n

Kunzy Nguyễn: Mik ko có ý chê bạn đâu nhưng mà câu a mik thấy bạn giải có chút gọi là ''sai''!

\(\lim\dfrac{\left(2n-1\right)\left(3n^2+2\right)^3}{-2n^5+4n^3-1}=\lim\dfrac{\left(\dfrac{2n-1}{n}\right)\left(\dfrac{3n^2+2}{n^2}\right)^3}{\dfrac{-2n^5+4n^3-1}{n^7}}\)

\(=\lim\dfrac{\left(2-\dfrac{1}{n}\right)\left(3+\dfrac{2}{n^2}\right)^3}{-\dfrac{2}{n^2}+\dfrac{4}{n^4}-\dfrac{1}{n^7}}=-\infty\)

\(\lim3^n\left(6.\left(\dfrac{2}{3}\right)^n-5+\dfrac{7n}{3^n}\right)=+\infty.\left(-5\right)=-\infty\)

a)

\(\left(\frac{1}{3}\right)^n\cdot27^n=3^n\)

\(\Rightarrow\left(\frac{1}{3}\cdot27\right)^n=3^n\)

\(\Rightarrow9^n=3^n\)

\(\Rightarrow\left(3^2\right)^n=3^n\)

\(\Rightarrow3^{2n}=3^n\)

\(\Rightarrow2n=n\)

\(\Leftrightarrow n=0\)

Vậy \(n=0\)

d) Ta có:

\(6^{3-n}=216\)

\(\Rightarrow6^{3-n}=6^3\)

\(\Rightarrow3-n=3\)

\(\Rightarrow n=3-3\)

\(\Rightarrow n=0\)

Vậy \(n=0\)\(\text{ }\)

e) Ta có:

\(5^n+5^{n+2}=650\)

\(\Rightarrow5^n+5^n\cdot5^2=650\)

\(\Rightarrow5^n\cdot\left(1+5^2\right)=650\)

\(\Rightarrow5^n\cdot26=25\cdot26\)

\(\Rightarrow5^n=25\)

\(\Rightarrow5^n=5^2\)

\(\Rightarrow n=2\)

Vậy \(n=2\)