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Ta có : \(3^{-1}.3^n+5.3^{n+1}=162\)

\(\Leftrightarrow3^{-1}.3^n+15.3^n=162\)

\(\Leftrightarrow3^n\left(3^{-1}+15\right)=162\)

\(\Leftrightarrow3^n\frac{46}{3}=162\)

\(\Leftrightarrow3^n=\frac{162.3}{46}=\frac{243}{23}\) (đề sai òi e ơi)

20 tháng 4 2018

không sai nhé ey

1 tháng 1 2017

\(3^{n-1}+9.3^n=28.3^5\)

\(\Rightarrow3^{n-1}+9.3^{n-1}.3=28.3^5\)

\(\Rightarrow3^{n-1}.\left(1+9.3\right)=28.3^5\)

\(\Rightarrow3^{n-1}.28=28.3^5\)

\(\Rightarrow3^{n-1}=3^5\)

\(\Rightarrow n-1=5\)

\(\Rightarrow n=6\)

Vậy n = 6

8 tháng 3 2017

mấy bài này để để mấy bạn khác làm,HUY TƯ là cộng tác viên nên làm những bài khó hơnoe

a: \(5^3\cdot25^n=5^{3n}\)

\(\Leftrightarrow5^{3n}=5^3\cdot5^{2n}\)

=>3n=2n+3

hay n=3

b: \(a^{\left(2n+6\right)\left(3n-9\right)}=1\)

=>(2n+6)(3n-9)=0

=>n=-3 hoặc n=3

c: \(\dfrac{1}{3}\cdot3^n=7\cdot3^2\cdot3^4-2\cdot3^n\)

\(\Leftrightarrow3^n\cdot\dfrac{1}{3}+3^n\cdot2=7\cdot3^6\)

\(\Leftrightarrow3^n=3^7\)

hay n=7

24 tháng 9 2017

a) \(\dfrac{1}{9}.27^n=3^n\)

\(\Leftrightarrow\dfrac{1}{9}=3^n:27^n\)

\(\Leftrightarrow\dfrac{1}{9}=\left(\dfrac{3}{27}\right)^n\)

\(\Leftrightarrow\dfrac{1}{9}=\left(\dfrac{1}{9}\right)^n\)

\(\Leftrightarrow n=1\)

b) \(3^{-2}.3^4.3^n=3^7\)

\(\Leftrightarrow3^2.3^n=3^7\)

\(\Leftrightarrow3^n=3^7:3^2\)

\(\Leftrightarrow3^n=3^5\)

\(\Leftrightarrow n=5\)

c) \(32^{-n}.16^n=2048\)

\(\Leftrightarrow\left(2^5\right)^{-n}.\left(2^4\right)^n=2^{11}\)

\(\Leftrightarrow2^{-5n}.2^{4n}=2^{11}\)

\(\Leftrightarrow2^{-n}=2^{11}\)

\(\Leftrightarrow n=-11\)

18 tháng 11 2019

Ta có

\(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)   và \(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n}-\frac{1}{n+1}-\frac{1}{n+2}\)  nên

\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{n\left(n+1\right)}+...+\frac{1}{2008\cdot2009}=1-\frac{1}{2009}=\frac{2008}{2009}\)

\(2B=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}+...+\frac{2}{2008\cdot2009\cdot2010}\)

\(=\frac{1}{1\cdot2}-\frac{1}{2009\cdot2010}=\frac{201944}{2009\cdot2010}\)

\(\Rightarrow B=\frac{1}{2}\cdot\frac{201944}{2009\cdot2010}=\frac{1009522}{2009\cdot2010}\)

Do đó \(\frac{B}{A}=\frac{1009522}{2009\cdot2010}:\frac{2008}{2009}=\frac{1009522\cdot2009}{2008\cdot2009\cdot2010}=\frac{5047611}{2018040}\)

16 tháng 7 2018

\(a)A=\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)

\(A=\dfrac{2^{12}.3^5-\left(2^2\right)^63.\left(3^2\right)^2}{\left(2^2\right)^6.3^6+\left(2^3\right)^4.3^5}-\dfrac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{\left(5^3\right)^3.7^3+5^9.\left(7.2\right)^3}\)

\(A=\dfrac{2^{12}.3^5-2^{12}.3^5}{2^{12}.3^6+2^{12}.3^5}-\dfrac{5^{10}.7^3-5^{10}.7^4}{5^6.7^3+5^9.7^3.2^3}\)

\(A=\dfrac{0}{2^{12}.3^6+2^{12}.3^5}-\dfrac{5^{10}.7^3\left(1-7\right)}{5^6.7^3\left(1+5^3+2^3\right)}\)

\(A=0-\dfrac{5^4.\left(-6\right)}{1+125+8}\)

\(A=0-\dfrac{625.\left(-6\right)}{134}\)

\(A=\dfrac{-3750}{134}\)\(=\dfrac{-1875}{67}\)

\(b)3^{n+2}-2^{n+2}+3^n-2^n\)

\(=3^n.3^2-2^n.2^2+3^n-2^n\)

\(=(3^n.9+3^n)-\left(2^n.4+2^n\right)\)

\(=3^n.10-2^n.5\)

\(=3^n.10-2^{n-1}.10\)

\(=10\left(3^n-2^{n-1}\right)⋮10\)

\(Suy\) \(ra:\) \(3^{n+2}-2^{n+2}+3^n-2^n⋮10\)

16 tháng 7 2018

b. Ta có: \(3^{n +2}-2^{n+2}+3^n-2^n\)

\(=\left(3^{n+2}+3^n\right)-\left(2^{n+2}+2^n\right)\)

\(=\left(3^n.3^2+3^n\right)-\left(2^{n-1}.2^3+2^{n-1}.2\right)\)

\(=3^n.\left(3^2+1\right)-2^{n-1}\left(2^3+2\right)\)

\(=3^n.10-2^{n-1}.10⋮10\)

17 tháng 2 2016

a)\(\frac{1}{3^2}\cdot3^{3n}=3^n\Rightarrow3=3^{3n-2}=3^n\Rightarrow3n-2=n\Rightarrow n=1\)

b)\(\frac{1}{3^2}\cdot3^4\cdot3^n=3^7\Rightarrow3^{n+2}=3^7\Rightarrow n+2=7\Rightarrow n=5\)