a/ \(\left(\frac{1}{3}\right)^m=\frac{1}{81}\)
\(\Leftrightarrow\left(\frac{1}{3}\right)^m=\left(\frac{1}{3}\right)^4\)
\(\Leftrightarrow m=4\left(tm\right)\)
b/ \(\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5\)
\(\Leftrightarrow n=10\)
\(\Leftrightarrow\left(\frac{3}{5}\right)^n=\left(\frac{3}{5}\right)^{10}\)
a)(1/3)^m=(1/3)^4
b)(3/5)^n=(3/5)^10
c)(-0,25)^p=(-0,25)^4
a)\(\left(\frac{1}{3}\right)^m=\frac{1}{81}\)
\(\left(\frac{1}{3}\right)^m=\left(\frac{1}{3}\right)^4\)
\(\Rightarrow m=4\)
vay \(m=4\)
b) \(\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5\)
\(\left(\frac{3}{5}\right)^n=\left(\frac{3}{5}\right)^{10}\)
\(\Rightarrow n=10\)
vay \(n=10\)
c) \(\left(-0,25\right)^p=\frac{1}{256}\)
\(\left(\frac{-1}{4}\right)^p=\left(\frac{1}{4}\right)^4\)
\(\Rightarrow p=4\)
voi mu duong thi \(\left(\frac{-1}{4}\right)^4=\left(\frac{1}{4}\right)^4\)
vay \(p=4\)
a, \(\frac{1}{81}=\left(\frac{1}{3}\right)^4\Rightarrow m=4\)
b, \(\left(\frac{9}{25}\right)^5=\left(\left(\frac{3}{5}\right)^2\right)^2\Rightarrow n=2.2=4\)
c, \(\frac{1}{256}=\left(\frac{1}{4}\right)^4\Rightarrow p=4\)
a) \(\left(\frac{1}{3}\right)^m=\frac{1}{81}\)
\(\Rightarrow\frac{1^m}{3^m}=\frac{1}{81}\Leftrightarrow\hept{\begin{cases}1^m=1\\3^m=81\end{cases}}\)
=> 81 = 3m = 34 <=> x = 4
b) \(\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5\)
\(\Rightarrow\frac{3^n}{5^n}=\frac{9^5}{25^5}\)
\(\frac{3^n}{5^n}=\frac{\left(3^2\right)^5}{\left(5^2\right)^5}\)
\(\frac{3^n}{5^n}=\frac{3^{10}}{5^{10}}\)
Vậy 3n = 310 tương đương với x = 10
5n = 510 tương đương với x = 10
<=> x = 10
c) \(\left(-0,25\right)^p=\frac{1}{256}\)
\(\left(\frac{-1}{4}\right)^p=\frac{1}{256}\)
\(\frac{\left(-1\right)^p}{4^p}=\frac{1}{256}\)
=> p là 1 số chẵn
=> 4p = 256
<=> 256 = 4p = 44
<=> x = 4
