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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/ \(\Rightarrow\frac{\left(-3\right)^n}{81}=-27\Rightarrow\left(-3\right)^n=-2187\Rightarrow\left(-3\right)^n=\left(-3\right)^7\Rightarrow n=7\)
b/ \(\Rightarrow-\frac{3}{8}-x+\frac{5}{6}=\frac{4}{3}\Rightarrow\frac{11}{24}-x=\frac{4}{3}\Rightarrow x=-\frac{7}{8}\)
Ta có :
\(B=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+...+\frac{1}{x}.\left(1+2+3+...+x\right)\)
\(B=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+...+\frac{1}{x}.\frac{x.\left(x+1\right)}{2}\)
\(B=1+\frac{3}{2}+\frac{4}{2}+...+\frac{x+1}{2}\)
\(B=\frac{2+3+4+...+\left(x+1\right)}{2}\)
để B = 115 thì \(\frac{2+3+4+...+\left(x+1\right)}{2}=115\)
\(\Rightarrow\)\(\left(x+3\right)x=115.2.2\)
\(\Rightarrow\)\(\left(x+3\right)x=23.20\)
\(\Rightarrow\)x = 20
\(a,\left[\left(0,5\right)^3\right]^n=\frac{1}{64}\Rightarrow\left(0,125\right)^n=0,125^2\Rightarrow n=2\)
\(b,\frac{64}{\left(-2\right)^{n+1}}=4\Rightarrow\left(-2\right)^{n+1}=\frac{64}{4}\Rightarrow\left(-2\right)^{n+1}=16\Rightarrow\left(-2\right)^{n+1}=\left(-2\right)^4\)
\(\Rightarrow n+1=4\Rightarrow n=3\)
\(c,\left(\frac{1}{3}\right)^{n+1}=\frac{1}{81}\Rightarrow\left(\frac{1}{3}\right)^{n+1}=\left(\frac{1}{3}\right)^4\Rightarrow n+1=4\Rightarrow n=3\)
\(d,\left(\frac{3}{4}\right)^n.\frac{1}{2}=\frac{81}{512}\Rightarrow\left(\frac{3}{4}\right)^n=\frac{81}{512}:\frac{1}{2}=\frac{81}{256}\Rightarrow\left(\frac{3}{4}\right)^n=\left(\frac{3}{4}\right)^4\Rightarrow n=4\)
Đề vậy phải ko bn (1/3)n=1/81
(1/3)n=1/81
(1/3)n=(1/3)4
==> n=4