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a) \(\frac{16}{2^n}=2\)
\(\Rightarrow16=2\cdot2^n\)
\(\Rightarrow16=2^{n+1}\)
\(\Rightarrow2^4=2^{n+1}\)
\(\Rightarrow n+1=4\)
\(\Rightarrow n=4-1\)
\(\Rightarrow n=3\)
Vậy n=3
b) \(\frac{\left(-3\right)^n}{81}=-27\)
\(\Rightarrow\left(-3\right)^n=-27\cdot81\)
\(\Rightarrow\left(-3\right)^n=\left(-3\right)^3.\left(-3\right)^4\)
\(\Rightarrow\left(-3\right)^n=\left(-3\right)^7\)
\(\Rightarrow n=7\)
Vậy n=7
c ) \(8^n:2^n=4\)
\(\Rightarrow2^{3n}:2^n=2^2\)
\(\Rightarrow2^{2n}=2^2\)
\(\Rightarrow2n=2\)
\(\Rightarrow n=2:2\)
\(\Rightarrow n=1\)
Vậy n=1

Ta có : \(\left(\dfrac{-10}{3}\right)^5.\left(\dfrac{-6}{5}\right)^4=\dfrac{\left(-10\right)^5.\left(-6\right)^4}{3^5.5^4}=\dfrac{\left(-2\right)^5.\left(-2\right)^4.5^5.3^4}{3^5.5^4}=\dfrac{\left(-2\right)^9.5}{3}\)

\(=\frac{2}{3}-\left[\left(\frac{-7}{4}\right)-\frac{1}{2}-\frac{3}{8}\right]=\frac{2}{3}+\frac{7}{4}+\frac{1}{2}+\frac{3}{8}=\frac{79}{24}\)
a) \(\dfrac{16}{2^n}=2\Rightarrow2^n=\dfrac{16}{2}=8=2^3\)
\(\Rightarrow n=3\)
b) \(\dfrac{\left(-3\right)^n}{81}=-27\)
\(\Leftrightarrow\dfrac{\left(-3\right)^n}{3^4}=\left(-3\right)^3\)
\(\Leftrightarrow\left(-3\right)^n=-\left(3^4\cdot3^3\right)=\left(-3\right)^7\)
\(\Rightarrow n=7\)
c) \(8^n:2^n=4\)
\(\Leftrightarrow2^{3n}:2^n=2^2\)
\(\Leftrightarrow3n-n=2\)
\(\Leftrightarrow2n=2\Rightarrow n=1\)
a, \(\dfrac{16}{2^n}=2\Rightarrow2^n.2=16\Rightarrow2^{n+1}=2^4\Rightarrow n+1=4\Rightarrow n=3\). Vậy n=3.
b, \(\dfrac{\left(-3\right)^n}{81}=-27\Rightarrow\left(-3\right)^n=81.\left(-27\right)\Rightarrow\left(-3\right)^n=-2187\)
\(\Rightarrow\left(-3\right)^n=\left(-3\right)^7\Rightarrow n=7\)
Vậy n=7
c, \(8^n:2^n=4\Rightarrow\left(2^3\right)^n:2^n=2^2\Rightarrow2^{3n}:2^n=2^2\Rightarrow2^{2n}=2^2\Rightarrow2n=2\Rightarrow n=1\)
Vậy n=1.