3ⁿ:9=27

=>3ⁿ:3²=3³

=>3ⁿ=3³.3²=3³⁺²=3⁵

=>n=5

(2.n+1³)=27

=>2.n+1=27

=>2.n=27-1=26

=>n=26:2

=>n=13

(-3)^n=(-3)^7

n=7

\(\dfrac{\left(-3\right)^n}{9^2}=\left(-27\right)\)

\(\Leftrightarrow\left(-3\right)^n=\left(-3\right)^7\)

hay n=7

len vietjack la co het

Bài 2: 

a) Ta có: \(A=\dfrac{4}{n-1}+\dfrac{6}{n-1}-\dfrac{3}{n-1}\)

\(=\dfrac{4+6-3}{n-1}\)

\(=\dfrac{7}{n-1}\)

Để A là số tự nhiên thì \(7⋮n-1\)

\(\Leftrightarrow n-1\inƯ\left(7\right)\)

\(\Leftrightarrow n-1\in\left\{1;7\right\}\)

hay \(n\in\left\{2;8\right\}\)

Vậy: \(n\in\left\{2;8\right\}\)

ta có B=2n+9/n+2-3n+5n+1/n+2=4n+10/n+2                                                   Để B là STN thì 4n+10⋮n+2                          4n+8+2⋮n+2                                  4n+8⋮n+2                                                      ⇒2⋮n+2                                     n+2∈Ư(2)                                                        Ư(2)={1;2}                                  Vậy n=0                                                                                  

Ta có: =4+6−3n−1=4+6−3�−1

\(\frac{1}{21}+\frac{1}{27}+\frac{1}{36}+...+\frac{2}{n\left(n+1\right)}=\frac{2}{9}\)

\(\frac{2}{42}+\frac{2}{54}+...+\frac{2}{n\left(n+1\right)}=\frac{2}{9}\)

\(\frac{2}{6.7}+\frac{2}{7.8}+...+\frac{2}{n\left(n+1\right)}=\frac{2}{9}\)

\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{1}{9}\)

\(\Rightarrow\frac{1}{6}-\frac{1}{n+1}=\frac{n+1-6}{6n+6}=\frac{1}{9}\)

\(\frac{n-5}{6n+6}=\frac{1}{9}\)

\(9n-45=6n+6\)

\(9n-6n=6+45=51\)

\(n=51:3=17\)

\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{n}.\left(n+1\right)=\frac{2}{9}\)

\(\Leftrightarrow\frac{1}{3.7}+\frac{1}{4.7}+\frac{1}{4.9}+...+\frac{2}{n}.\left(n+1\right)=\frac{2}{9}\)

\(\Leftrightarrow\frac{2}{2.3.7}+\frac{2}{2.4.7}+\frac{2}{2.4.9}+...+\frac{2}{n}.\left(n+1\right)=\frac{2}{9}\)

\(\Leftrightarrow\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+...+\frac{2}{n}.\left(n+1\right)=\frac{2}{9}\)

\(\Leftrightarrow2.\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{n}-\frac{1}{n}+1\right)=\frac{2}{9}\)

\(\Leftrightarrow2.\left(\frac{1}{6}-\frac{1}{n}+1\right)=\frac{2}{9}\)

\(\Leftrightarrow\frac{1}{6}-\frac{1}{n}+1=\frac{1}{9}\)

\(\Leftrightarrow\frac{1}{n}+1=\frac{1}{6}-\frac{1}{9}\)

\(\Leftrightarrow\frac{1}{n}+1=\frac{1}{18}\)

\(\Leftrightarrow n+1=18\)

\(\Leftrightarrow n=17\)

Vậy \(n=17\)

\(27^n.9^n=9^{27}:81\)

\(3^{3n}.3^{2n}=3^{54}:3^4\)

\(3^{5n}=3^{50}\)

=> 5n = 50

=> n = 10

b: =>\(\dfrac{2}{2}+\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{n\left(n+1\right)}=\dfrac{200}{101}\)

=>\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{n\left(n+1\right)}=\dfrac{100}{101}\)

=>1-1/2+1/2-1/3+...+1/n-1/n+1=100/101

=>1-1/(n+1)=100/101

=>1/(n+1)=1/101

=>n+1=101

=>n=100

mk ghi lại đề nha:

27ⁿ : 9ⁿ = 9²⁷ : 81

(27 : 9)ⁿ = 9²⁷ : 9²

\(\Rightarrow\)     3ⁿ  = 9²⁵

\(\Rightarrow\)      3ⁿ = (3²)²⁵

\(\Rightarrow\)      3ⁿ  = 3⁵⁰

Vậy n = 50

\(27^n.9^n=9^{27}:81\Rightarrow3^{3n}:3^{2n}=3^{54}:3^4=3^{50}\)

\(\Rightarrow3^{5n}=3^{50}\Rightarrow5n=50\Rightarrow n=\frac{50}{5}=10\)

\(27^n.9^n=9^{27}:81\)

\(\left(27.9\right)^n=9^{27}:9^2\)

\(\left(27.9\right)^n=9^{25}\)

\(\left(3^3.3^2\right)^n=9^{25}\)

\(\left(3^5\right)^n=\left(3^2\right)^{25}\)

\(\left(3^5\right)^n=\left(3^5\right)^{10}\)

\(\Rightarrow n=10\)

Bài 6 :

a) \(\dfrac{625}{5^n}=5\Rightarrow\dfrac{5^4}{5^n}=5\Rightarrow5^{4-n}=5^1\Rightarrow4-n=1\Rightarrow n=3\)

b) \(\dfrac{\left(-3\right)^n}{27}=-9\Rightarrow\dfrac{\left(-3\right)^n}{\left(-3\right)^3}=\left(-3\right)^2\Rightarrow\left(-3\right)^{n-3}=\left(-3\right)^2\Rightarrow n-3=2\Rightarrow n=5\)

c) \(3^n.2^n=36\Rightarrow\left(2.3\right)^n=6^2\Rightarrow\left(6\right)^n=6^2\Rightarrow n=6\)

d) \(25^{2n}:5^n=125^2\Rightarrow\left(5^2\right)^{2n}:5^n=\left(5^3\right)^2\Rightarrow5^{4n}:5^n=5^6\Rightarrow\Rightarrow5^{3n}=5^6\Rightarrow3n=6\Rightarrow n=3\)

Bài 7 :

a) \(3^x+3^{x+2}=9^{17}+27^{12}\)

\(\Rightarrow3^x\left(1+3^2\right)=\left(3^2\right)^{17}+\left(3^3\right)^{12}\)

\(\Rightarrow10.3^x=3^{34}+3^{36}\)

\(\Rightarrow10.3^x=3^{34}\left(1+3^2\right)=10.3^{34}\)

\(\Rightarrow3^x=3^{34}\Rightarrow x=34\)

b) \(5^{x+1}-5^x=100.25^{29}\Rightarrow5^x\left(5-1\right)=4.5^2.\left(5^2\right)^{29}\)

\(\Rightarrow4.5^x=4.25^{2.29+2}=4.5^{60}\)

\(\Rightarrow5^x=5^{60}\Rightarrow x=60\)

c) Bài C bạn xem lại đề

d) \(\dfrac{3}{2.4^x}+\dfrac{5}{3.4^{x+2}}=\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{10}}\)

\(\Rightarrow\dfrac{3}{2.4^x}-\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{x+2}}-\dfrac{5}{3.4^{10}}=0\)

\(\Rightarrow\dfrac{3}{2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)+\dfrac{5}{3.4^2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)=0\)

\(\Rightarrow\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)\left(\dfrac{3}{2}+\dfrac{5}{3.4^2}\right)=0\)

\(\Rightarrow\dfrac{1}{4^x}-\dfrac{1}{4^8}=0\)

\(\Rightarrow\dfrac{4^8-4^x}{4^{x+8}}=0\Rightarrow4^8-4^x=0\left(4^{x+8}>0\right)\Rightarrow4^x=4^8\Rightarrow x=8\)

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a) 2n.16=128

=>32n=128

=>n=128:32

=>n=4

b)3n.9=27

=>27n=27

=>n=27:27

=>n=1

c)(2n+1)³=27

=>(2n+1)³=3³

=>2n+1=3

=>2n=3-1

=>2n=2

=>n=2:2

=>n=1