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2n-1:2=256
2n-1=512=29=>n-1=9=>n=10
5n+5n-2=650
5n-2(25+1)=650=>5n-2=25=52
=>n-2=2=>n=4
\(\dfrac{625}{5^n}\)=5
=>\(\dfrac{5^4}{5^n}\) =5
=>\(5^4\) :\(5^n\) = 5
=>\(5^{4-n}\) =\(5^1\)
=>4\(-\)n=1
=>n=4-1
=>n=3
a)
\(\left(\frac{1}{3}\right)^n\cdot27^n=3^n\)
\(\Rightarrow\left(\frac{1}{3}\cdot27\right)^n=3^n\)
\(\Rightarrow9^n=3^n\)
\(\Rightarrow\left(3^2\right)^n=3^n\)
\(\Rightarrow3^{2n}=3^n\)
\(\Rightarrow2n=n\)
\(\Leftrightarrow n=0\)
Vậy \(n=0\)
d) Ta có:
\(6^{3-n}=216\)
\(\Rightarrow6^{3-n}=6^3\)
\(\Rightarrow3-n=3\)
\(\Rightarrow n=3-3\)
\(\Rightarrow n=0\)
Vậy \(n=0\)\(\text{ }\)
a) \(\left(\frac{1}{3}\right)^n=\frac{1}{27}\)
\(\left(\frac{1}{3}\right)^n=\left(\frac{1}{3}\right)^3\)
\(\Rightarrow n=3\)
b) \(\left(\frac{3}{5}\right)^n=\frac{81}{625}\)
\(\left(\frac{3}{5}\right)^n=\left(\frac{3}{5}\right)^4\)
\(\Rightarrow n=4\)
c) \(3^n\cdot2^n=36\)
\(\left(3\cdot2\right)^n=36\)
\(6^n=6^2\)
\(\Rightarrow n=2\)
d) \(\frac{2^n}{3^n}=\frac{8}{27}\)
\(\left(\frac{2}{3}\right)^n=\left(\frac{2}{3}\right)^3\)
\(\Rightarrow n=3\)
1. Ta có: \(x\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)
=> \(x\left(6-x\right)^{2003}-\left(6-x\right)^{2003}=0\)
=> \(\left(6-x\right)^{2003}\left(x-1\right)=0\)
=> \(\orbr{\begin{cases}\left(6-x\right)^{2003}=0\\x-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}6-x=0\\x=1\end{cases}}\)
=> \(\orbr{\begin{cases}x=6\\x=1\end{cases}}\)
Bài 2. Ta có: (3x - 5)100 \(\ge\)0 \(\forall\)x
(2y + 1)100 \(\ge\)0 \(\forall\)y
=> (3x - 5)100 + (2y + 1)100 \(\ge\)0 \(\forall\)x;y
Dấu "=" xảy ra khi: \(\hept{\begin{cases}3x-5=0\\2y+1=0\end{cases}}\) => \(\hept{\begin{cases}3x=5\\2y=-1\end{cases}}\) => \(\hept{\begin{cases}x=\frac{5}{3}\\y=-\frac{1}{2}\end{cases}}\)
Vậy ...
a)
\(2^{n-1}:2=256\)
\(\Rightarrow2^{n-1}:2=2^8\)
\(\Rightarrow2^{n-1}=2^9\)
\(\Rightarrow n-1=9\)
\(\Rightarrow n=10\)
b)
\(5^n+5^{n-2}=650\)
\(\Rightarrow5^n+5^n:5^2=650\)
\(\Rightarrow5^n+5^n:25=650\)
\(\Rightarrow5^n+5^n.\dfrac{1}{25}=650\)
\(\Rightarrow5^n.\left(1+\dfrac{1}{25}\right)=650\)
\(\Rightarrow5^n.\dfrac{26}{25}=650\)
\(\Rightarrow5^n=625\)
\(\Rightarrow5^n=5^4\)
\(\Rightarrow n=4\)
c)
\(2^{n-3}+2^{n+1}=136\)
\(\Rightarrow2^n.\dfrac{1}{2^3}+2^n.2=136\)
\(\Rightarrow2^n.\left(\dfrac{1}{8}+2\right)=136\)
\(\Rightarrow2^n.\dfrac{17}{8}=136\)
\(\Rightarrow2^n=64\)
\(\Rightarrow2^n=2^6\)
\(\Rightarrow n=6\)
Thế mà "T lên chỉ để kiếm đề thôi", ngứa **t