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a: \(\Leftrightarrow3^n:27^n=\dfrac{1}{9}\)
\(\Leftrightarrow\left(\dfrac{1}{9}\right)^n=\dfrac{1}{9}\)
hay n=1
b: \(\Leftrightarrow3^n\cdot3^2=3^8\)
=>n+2=8
hay n=6
c: \(\Leftrightarrow2^n\cdot\dfrac{9}{2}=9\cdot2^5\)
\(\Leftrightarrow2^n=2^6\)
hay n=6
d: \(\Leftrightarrow8^n=512\)
hay n=3
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)\(3^{-2}.3^4.3^n=3^7\)
\(\Leftrightarrow3^{2+n}=3^7\)
\(\Leftrightarrow2+n=7\)
\(\Leftrightarrow n=5\)
Bạn tham khảo tại đây nhé: Câu hỏi của Khánh Huyền⁀ᶦᵈᵒᶫ .
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a, \(\frac{1}{9}.27^n=3^n\Leftrightarrow\frac{1}{9}.3^{3.n}=3^n\Leftrightarrow\frac{1}{3^2}=3^n:3^{3n}\Leftrightarrow\frac{1}{3^2}=3^{n-3n}=3^{2n}\)
=> 3^2n . 3^2 = 1 => 3^( 2n + 2) = 3^0 => 2n + 2 = 0 => 2n = - 2 => n = - 1
b, 3^-2.3^4 .3^n = 3^ 7 => 3^ ( -2 + 4 + n) = 3^7 => 3^ (n+ 2) = 3^7 => n + 2 = 7 => n = 5
\(A=1+3+3^2+3^3+...+3^{101}\)
\(3A=3+3^2+3^3+3^4+...+3^{101}\)
\(3A-A=\left(3+3^2+3^3+3^4+...+3^{101}\right)-\left(1+3+3^2+3^3+...+3^{100}\right)\)
\(2A=3^{101}-1\)
\(A=\left(3^{101}-1\right):2\)
Thu gọn tổng sau:
A=1+3+32+33+...+3100
B= 2100-299-298-297-...-22-2
C= 3100-399+398-397-...+32-3+1
a)\(32^{-n}\cdot16^n=2048\)
\(\left(2^5\right)^{-n}\cdot\left(2^4\right)^n\)=2048
\(2^{-5n}\cdot2^{4n}\)=\(2^{11}\)
\(2^{-5n+4n}=2^{11}\)
\(2^{-x}=2^{11}\)
\(\Rightarrow x=-11\)
b)\(2^{-1}\cdot2^n+4\cdot2^n=9\cdot2^5\)
\(\frac{1}{2}\cdot2^n+4\cdot2^n=288\)
\(2^n\left(\frac{1}{2}+4\right)=288\)
\(2^n\cdot\frac{9}{2}=288\)
\(2^n=288:\frac{9}{2}\)
\(2^n=64\)
\(2^n=2^6\)
\(\Rightarrow n=6\)
a) 32-n . 16n = 2048
\(\frac{1}{32n}\) . 16n = 2048
\(\frac{1}{2^n.16^n}\) . 16n = 2048
\(\frac{1}{2^n}\) = 2048
2-n = 2048
2-n = 211
\(\Rightarrow\) -n = 11
\(\Rightarrow\) n = -11
Vậy n = -11
a) \(\dfrac{1}{9}.27^n=3^n\)
\(\dfrac{1}{3^2}.3^{3n}=3^n\\ \Rightarrow3^{3n-2}=3^n\\ \Rightarrow3n-2=n\\ \Rightarrow n=1\)
b) \(3^{-2}.3^4.3^n=3^7\)
\(\dfrac{1}{3^2}.3^4.3^n=3^7\\ \Rightarrow3^{n+2}=3^7\Rightarrow n+2=7\\ \Rightarrow n=5\)
c) \(2^{-1}.2^n+4.2^n=9.2^5\)
\(\dfrac{1}{2}.2^n+4.2^n=9.2^5\\ \Rightarrow2^n\left(\dfrac{1}{2}+4\right)=9.2^5\\ \Rightarrow2^{n-1}.9=9.2^5\\ \Rightarrow n-1=5\\ \Rightarrow n=6\)
d) \(32^{-n}.16^{-n}=2048\)
\(\dfrac{1}{2^n.16^n}.16^n=2^{11}=\dfrac{1}{2^n}=2^{11}\\ \Rightarrow2^n.2^{11}=1\\ \Rightarrow2^{n+11}=2^0\\ \Rightarrow n+11=0\\ \Rightarrow n=-11\)
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cảm ơn bạn nhá! tặng bn nè