Ai giải giúp với :
Tìm số nguyên n biết
a) 1/9 . 27^n = 3^n
b) 3^ -2 . 3^4 . 3^n = 37
c) 32^ -n . 16^n = 2048
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a) \(\frac{1}{9}.27^n=3^n\)
\(\Leftrightarrow3^{-2}.3^{3n}=3^n\)
\(\Leftrightarrow3^{3n-2}=3^n\)
\(\Leftrightarrow3n-2=n\)
\(\Leftrightarrow2n=2\)
\(\Leftrightarrow n=1\)
b)\(3^{-2}.3^4.3^n=3^7\)
\(\Leftrightarrow3^{2+n}=3^7\)
\(\Leftrightarrow2+n=7\)
\(\Leftrightarrow n=5\)
a)
\(\Rightarrow3^{-2}.\left(3^3\right)^n=3^n\)
\(\Rightarrow3^{-2}.3^{3n}=3^n\)
\(\Rightarrow3^{3n-2}=3^n\)
\(\Rightarrow3n-2=n\)
\(\Rightarrow n=1\)
b)
\(\Rightarrow3^{4+n-2}=3^7\)
\(\Rightarrow x^{n+2}=3^7\)
\(\Rightarrow n+2=7\)
\(\Rightarrow n=5\)
c)
\(\Rightarrow2^n\left(\frac{1}{2}+4\right)=9.2^5\)
\(\Rightarrow2^n.4,5=9.2^5\)
\(\Rightarrow2^n=2.2^5\)
\(\Rightarrow2^n=2^6\)
\(\Rightarrow n=6\)
d)
\(\Rightarrow\left(2^5\right)^{-n}.\left(2^4\right)^n=2048\)
\(\Rightarrow2^{n-5n}=2^{11}\)
\(\Rightarrow-4n=11\)
\(\Rightarrow n=-\frac{4}{11}\)
\(\frac{1}{9}.27^n=3^n\)
\(\Rightarrow\frac{3^n}{27^n}=\frac{1}{9}\)
\(\Rightarrow\left(\frac{3}{27}\right)^n=\frac{1}{9}\)
\(\Rightarrow\left(\frac{1}{9}\right)^n=\frac{1}{9}\)
\(\Rightarrow n=1\)
a) \(n^{51}=n\)
\(\Rightarrow n^{51}-n=0\)
\(n\left(n^{50}-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}n=0\\n^{50}-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}n=0\\n^{50}=1\end{cases}}\)
\(\Rightarrow n\in\left\{-1;0;1\right\}\)
b) \(\frac{1}{9}.27^n=3^n\)
\(\Rightarrow3^{-2}.3^{3n}=3^n\)
\(3^{3n-2}=3^n\)
\(\Rightarrow3n-2=n\)
\(3n-n=2\)
\(2n=2\)
\(n=2:2=1\)
c) \(3^{-2}.3^4.3^n=3^7\)
\(3^{n+4-2}=3^7\)
\(3^{n+2}=3^7\)
\(\Rightarrow n+2=7\)
\(\Rightarrow n-7=5\)
d) \(32^{-n}.16^n=2048\)
\(2^{-5n}.2^{4n}=2^{10}\)
\(2^{4n-5n}=2^{10}\)
\(2^{-n}=2^{10}\)
\(\Rightarrow-n=10\)
\(\Rightarrow n=-10\)
a)1/9.27^n=3^n
3^n=3^n
=>n={0;1;2;3...}
Tích nha ^_^ !!!