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\(\left(\frac{1}{2}\right)^n=\left(\frac{1}{8}\right)^5\)
\(\left(\frac{1}{2}\right)^n=\left(\frac{1^3}{2^3}\right)^5\)
\(\left(\frac{1}{2}\right)^n=\left[\left(\frac{1}{2}\right)^3\right]^5\)
\(\left(\frac{1}{2}\right)^n=\left(\frac{1}{2}\right)^{15}\)
n = 15
\(\left(\frac{1}{2}\right)^n=\left(\frac{1}{8}\right)^5\)
\(\Rightarrow\left(\frac{1}{2}\right)^n=\left(\frac{1}{2}\right)^{3.5}\)
\(\Rightarrow\left(\frac{1}{2}\right)^n=\left(\frac{1}{2}\right)^{15}\)
\(\Rightarrow n=15\)
Vậy n = 15
M = 1/3 + (1/3)^2 + ..+ (1/3)^2013
3M = 1 + 1/3+ ...+ (1/3)^2012
3M - M = 1+ 1/3 + (1/3)^2 + ..+ (1/3)^2012 - 1/3 - ...- (1/3)^2013
2M = (1/3 - 1/3) +..+[(1/3)^2012 -(1/3)2012]+ [1 - (1/3)^2013]
2M = 0 + 0 + .. + 0 + 1 - (1/3)^2013
2M = 1 - (1/3)^2013
1 - 2M = 1 - 1 + (1/3)^2013
1 - 2M = (1/3)^2013
(1/3)^2013 = (1/3)^n
2013 = n
Vậy n = 2013
M = 1/3 + (1/3)^2 + ..+ (1/3)^2013
3M = 1 + 1/3+ ...+ (1/3)^2012
3M - M = 1+ 1/3 + (1/3)^2 + ..+ (1/3)^2012 - 1/3 - ...- (1/3)^2013
2M = (1/3 - 1/3) +..+[(1/3)^2012 -(1/3)2012]+ [1 - (1/3)^2013]
2M = 0 + 0 + .. + 0 + 1 - (1/3)^2013
2M = 1 - (1/3)^2013
1 - 2M = 1 - 1 + (1/3)^2013
1 - 2M = (1/3)^2013
(1/3)^2013 = (1/3)^n
2013 = n
Vậy n = 2013
\(a,\left[\left(0,5\right)^3\right]^n=\frac{1}{64}\Rightarrow\left(0,125\right)^n=0,125^2\Rightarrow n=2\)
\(b,\frac{64}{\left(-2\right)^{n+1}}=4\Rightarrow\left(-2\right)^{n+1}=\frac{64}{4}\Rightarrow\left(-2\right)^{n+1}=16\Rightarrow\left(-2\right)^{n+1}=\left(-2\right)^4\)
\(\Rightarrow n+1=4\Rightarrow n=3\)
\(c,\left(\frac{1}{3}\right)^{n+1}=\frac{1}{81}\Rightarrow\left(\frac{1}{3}\right)^{n+1}=\left(\frac{1}{3}\right)^4\Rightarrow n+1=4\Rightarrow n=3\)
\(d,\left(\frac{3}{4}\right)^n.\frac{1}{2}=\frac{81}{512}\Rightarrow\left(\frac{3}{4}\right)^n=\frac{81}{512}:\frac{1}{2}=\frac{81}{256}\Rightarrow\left(\frac{3}{4}\right)^n=\left(\frac{3}{4}\right)^4\Rightarrow n=4\)
M = 1/3 + (1/3)^2 + ..+ (1/3)^2013
3M = 1 + 1/3+ ...+ (1/3)^2012
3M - M = 1+ 1/3 + (1/3)^2 + ..+ (1/3)^2012 - 1/3 - ...- (1/3)^2013
2M = (1/3 - 1/3) +..+[(1/3)^2012 -(1/3)2012]+ [1 - (1/3)^2013]
2M = 0 + 0 + .. + 0 + 1 - (1/3)^2013
2M = 1 - (1/3)^2013
1 - 2M = 1 - 1 + (1/3)^2013
1 - 2M = (1/3)^2013
(1/3)^2013 = (1/3)^n
2013 = n
Vậy n = 2013
M = 1/3 + (1/3)^2 + ..+ (1/3)^2013
3M = 1 + 1/3+ ...+ (1/3)^2012
3M - M = 1+ 1/3 + (1/3)^2 + ..+ (1/3)^2012 - 1/3 - ...- (1/3)^2013
2M = (1/3 - 1/3) +..+[(1/3)^2012 -(1/3)2012]+ [1 - (1/3)^2013]
2M = 0 + 0 + .. + 0 + 1 - (1/3)^2013
2M = 1 - (1/3)^2013
1 - 2M = 1 - 1 + (1/3)^2013
1 - 2M = (1/3)^2013
(1/3)^2013 = (1/3)^n
2013 = n
Vậy n = 2013
M = 1/3 + (1/3)^2 + ..+ (1/3)^2013
3M = 1 + 1/3+ ...+ (1/3)^2012
3M - M = 1+ 1/3 + (1/3)^2 + ..+ (1/3)^2012 - 1/3 - ...- (1/3)^2013
2M = (1/3 - 1/3) +..+[(1/3)^2012 -(1/3)2012]+ [1 - (1/3)^2013]
2M = 0 + 0 + .. + 0 + 1 - (1/3)^2013
2M = 1 - (1/3)^2013
1 - 2M = 1 - 1 + (1/3)^2013
1 - 2M = (1/3)^2013
(1/3)^2013 = (1/3)^n
2013 = n
Vậy n = 2013
4. (3/4-81)(3^2/5-81)(3^3/6-81)....(3^6/9-81).....(3^2011/2014-81)
mà 3^6/9-81=0 => (3/4-81)(3^2/5-81)....(3^2011/2014-81)=0
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