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a ) \(\left(\frac{1}{3}\right)^m=\left(\frac{1}{3}\right)^4\)
\(\Rightarrow m=4\)
b ) \(\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5\)
\(\Leftrightarrow\left(\frac{3}{5}^2\right)^n=\left(\frac{9}{25}\right)^5\)
\(\Leftrightarrow\left(\frac{9}{25}\right)^n=\left(\frac{9}{25}\right)^5\)
\(\Leftrightarrow n=5\)
c ) \(\left(-0,25\right)^p=\frac{1}{256}\)
\(\Leftrightarrow\left(-\frac{1}{4}\right)^p=\frac{1}{256}\)
\(\Leftrightarrow\left(-\frac{1}{4}\right)^p=\left(-\frac{1}{4}\right)^4\)
\(\Leftrightarrow p=4\)
\(a.\)
\(\left(\frac{1}{3}\right)^m=\frac{1}{81}\)
\(\Rightarrow\left(\frac{1}{3}\right)^m=\left(\frac{1}{3}\right)^4\)
\(\Rightarrow m=4\)
Vậy : \(m=4\)
\(b.\)
\(\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5\)
\(\Rightarrow\left(\frac{3}{5}\right)^n=\left(\frac{3}{5}\right)^{15}\)
\(\Rightarrow n=5\)
Vậy : \(n=5\)
\(c.\)
\(\left(-0,25\right)^p=\frac{1}{256}\)
\(\Rightarrow\left(-\frac{1}{4}\right)^p=\frac{1}{256}\)
\(\Rightarrow\left(-\frac{1}{4}\right)^p=\left(\frac{1}{4}\right)^4\)
\(\Rightarrow p=4\)
Vậy : \(p=4\)
a,32
b,\(-\frac{1}{10}\)
c,-1000000
d,\(\frac{9}{16}\)
a) \(\left(\frac{2}{5}-\frac{1}{2}\right)^2-\frac{11}{5}:\frac{-11}{5}=\left(-\frac{1}{10}\right)^2+1=1\frac{1}{100}\)
b) \(\left(-\frac{5}{7}\right)^2+8.\left(0,5\right)^2+\left(-1\right)^{2010}=\frac{25}{49}+2+1=3\frac{25}{49}\)
c) \(\frac{9999^2}{3333^2}+\left(0,5\right)^2.\left(-2\right)^4-\left(-\frac{4}{3}\right)^2=9+1-\frac{16}{9}=8\frac{2}{9}\)
d) \(\left|-\frac{2}{5}+\frac{1}{7}\right|:\frac{-3}{35}+\frac{-3}{7}.\frac{7}{5}=\frac{9}{35}.\frac{35}{-3}-\frac{3}{5}=-3\frac{3}{5}\)
e) \(\frac{1}{2}-\left(-0,4\right)+\frac{1}{3}+\frac{1}{5}-\frac{-1}{6}+\frac{-4}{35}+\frac{1}{41}\)
\(=\frac{1}{2}+\frac{2}{5}+\frac{1}{3}+\frac{1}{5}+\frac{1}{6}-\frac{4}{35}+\frac{1}{41}=1\frac{732}{1435}\)
a) \(\left(\frac{1}{3}\right)^n=\frac{1}{81}\)
\(\Rightarrow\left(\frac{1}{3}\right)^n=\frac{1^4}{3^4}\)
\(\Rightarrow\left(\frac{1}{3}\right)^n=\left(\frac{1}{3}\right)^4\)
\(\Rightarrow n=4\)
Vậy n = 4
b) \(\frac{-512}{343}=\left(\frac{-8}{7}\right)^n\)
\(\Rightarrow\frac{-8^3}{7^3}=\left(\frac{-8}{7}\right)^n\)
\(\Rightarrow\left(\frac{-8}{7}\right)^3=\left(\frac{-8}{7}\right)^n\)
\(\Rightarrow n=3\)
Vậy n = 3
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