Question

[piaiu]

tìm x,n

1.(-1/2.x+0.9)^4=16/81

2.(4/3)^n-2=16/9

3.(1/2)^2n-1=1/8

4.(-4/5)^3n+1=16/25

Answer 1

[456]

\(1.\left(-\dfrac{1}{2}\cdot x+0,9\right)^4=\dfrac{16}{81}\)

\(\Rightarrow\left(-\dfrac{1}{2}\cdot x+0,9\right)^4=\left(\pm\dfrac{2}{3}\right)^4\)

\(\Rightarrow\left[{}\begin{matrix}-\dfrac{1}{2}\cdot x+0,9=\dfrac{2}{3}\\-\dfrac{1}{2}\cdot x+0,9=-\dfrac{2}{3}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}-\dfrac{1}{2}\cdot x=\dfrac{2}{3}-0,9\\-\dfrac{1}{2}\cdot x=-\dfrac{2}{3}-0,9\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}-\dfrac{1}{2}\cdot x=-\dfrac{7}{30}\\-\dfrac{1}{2}\cdot x=-\dfrac{47}{30}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{7}{30}:-\dfrac{1}{2}\\x=-\dfrac{47}{30}:-\dfrac{1}{2}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{15}\\x=\dfrac{47}{15}\end{matrix}\right.\)

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 \(n=\dfrac{1}{3}\)