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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) \(32^{-n}.16^n=2048\)
\(\Leftrightarrow\left(2^5\right)^{-n}.\left(2^4\right)^n=2^{11}\)
\(\Leftrightarrow2^{-5n}.2^{4n}=2^{11}\)
\(\Leftrightarrow2^{-n}=2^{11}\)
\(\Leftrightarrow n=-11\)
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
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 \(\frac{B}{A}=\frac{1009522}{2009\cdot2010}:\frac{2008}{2009}=\frac{1009522\cdot2009}{2008\cdot2009\cdot2010}=\frac{5047611}{2018040}\)
\(3^{n-1}+9.3^n=28.3^5\)
\(\Rightarrow3^{n-1}+9.3^{n-1}.3=28.3^5\)
\(\Rightarrow3^{n-1}.\left(1+9.3\right)=28.3^5\)
\(\Rightarrow3^{n-1}.28=28.3^5\)
\(\Rightarrow3^{n-1}=3^5\)
\(\Rightarrow n-1=5\)
\(\Rightarrow n=6\)
Vậy n = 6
mấy bài này để để mấy bạn khác làm,HUY TƯ là cộng tác viên nên làm những bài khó hơn
a)\(\frac{1}{3^2}\cdot3^{3n}=3^n\Rightarrow3=3^{3n-2}=3^n\Rightarrow3n-2=n\Rightarrow n=1\)
b)\(\frac{1}{3^2}\cdot3^4\cdot3^n=3^7\Rightarrow3^{n+2}=3^7\Rightarrow n+2=7\Rightarrow n=5\)
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)
không sai nhé ey