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Ta có :\(\frac{1}{27}=\frac{81^n}{3^n}\)
\(\frac{1}{27}=81^n:3^n=27^n\)
\(27^n=27^{-1}\)
n = -1
Vậy n = -1
1/27 x 81^n =3^n
=>(1/3)^3 x (3^4)^n =3^n
=>(3^ -1)^3 x 3^4n = 3^n
=>3^-3 x 3^4n=3^n
=>3^(-3+4n)=3^n
=> -3+4n=n
=> 4n-n=3
=> 3n =3=>n=1
\(\frac{\left(-3\right)^x}{81}=-27\)
=> ( - 3 )x = - 2187
=> ( - 3 )x = ( - 3 )7
=> x = 7
\(\frac{27^x}{3^x}=81\)
\(\Rightarrow\frac{\left(3^3\right)^x}{3^x}=81\)
\(\Rightarrow\frac{3^{3x}}{3^x}=81\)
\(\Rightarrow3^{3x-x}=81\)
\(\Rightarrow3^{2x}=81\)
\(\Rightarrow3^{2x}=3^4\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=4:2\)
\(\Rightarrow x=2\)
Vậy \(x=2.\)
Chúc bạn học tốt!
a) \(\frac{75^3.3^7}{81^4.5^6}=\frac{5^3.3^3.5^3.3^7}{\left(3^4\right)^4.5^6}=\frac{5^6.3^3.3^7}{3^{16}.5^6}=\frac{3^{10}}{3^{16}}=\frac{1}{3^6}=\frac{1}{729}\)
b) \(\frac{6^6.4^2}{3^{12}.2^8}=\frac{2^6.3^6.\left(2^2\right)^2}{3^{12}.2^8}=\frac{2^6.3^6.2^4}{3^{12}.2^8}=\frac{2^{10}.3^6}{3^{12}.2^8}=\frac{2^2.1}{3^6}=\frac{4}{729}\)
c) \(\frac{34^5.2^5}{2^{14}.17^5}=\frac{2^5.17^5.2^5}{2^{14}.17^5}=\frac{2^{10}}{2^{14}}=\frac{1}{2^4}=\frac{1}{16}\)
\(\dfrac{1}{27}\cdot3^x=81\)
\(\Rightarrow3^x=81:\dfrac{1}{27}\)
\(\Rightarrow3^x=2187\)
\(\Rightarrow3^x=3^7\)
\(\Rightarrow x=7\)