\(1.\left(-\dfrac{1}{2}\cdot x+0,9\right)^4=\dfrac{16}{81}\)
\(\Rightarrow\left(-\dfrac{1}{2}\cdot x+0,9\right)^4=\left(\pm\dfrac{2}{3}\right)^4\)
\(\Rightarrow\left[{}\begin{matrix}-\dfrac{1}{2}\cdot x+0,9=\dfrac{2}{3}\\-\dfrac{1}{2}\cdot x+0,9=-\dfrac{2}{3}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}-\dfrac{1}{2}\cdot x=\dfrac{2}{3}-0,9\\-\dfrac{1}{2}\cdot x=-\dfrac{2}{3}-0,9\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}-\dfrac{1}{2}\cdot x=-\dfrac{7}{30}\\-\dfrac{1}{2}\cdot x=-\dfrac{47}{30}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{7}{30}:-\dfrac{1}{2}\\x=-\dfrac{47}{30}:-\dfrac{1}{2}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{15}\\x=\dfrac{47}{15}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{7}{15};\dfrac{47}{15}\right\}\)
\(2.\left(\dfrac{4}{3}\right)^{n-2}=\dfrac{16}{9}\)
\(\Rightarrow\left(\dfrac{4}{3}\right)^{n-2}=\left(\dfrac{4}{3}\right)^2\)
\(\Rightarrow n-2=2\)
\(\Rightarrow n=2+2\)
\(\Rightarrow n=4\)
Vậy \(n=4\)
\(3.\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{2n-1}=\left(\dfrac{1}{2}\right)^3\)
\(\Rightarrow2n-1=3\)
\(\Rightarrow2n=3+1\)
\(\Rightarrow2n=4\)
\(\Rightarrow n=4:2\)
\(\Rightarrow n=2\)
Vậy \(n=2\)
\(4.\left(-\dfrac{4}{5}\right)^{3n+1}=\dfrac{16}{25}\)
\(\Rightarrow\left(-\dfrac{4}{5}\right)^{3n+1}=\left(-\dfrac{4}{5}\right)^2\)
\(\Rightarrow3n+1=2\)
\(\Rightarrow3n=2-1\)
\(\Rightarrow3n=1\)
\(\Rightarrow n=1:3\)
\(\Rightarrow n=\dfrac{1}{3}\)
Vậy \(n=\dfrac{1}{3}\)
