Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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)\(3^{-2}.3^4.3^n=3^7\)
\(\Leftrightarrow3^{2+n}=3^7\)
\(\Leftrightarrow2+n=7\)
\(\Leftrightarrow n=5\)
\(A=1+3+3^2+3^3+...+3^{101}\)
\(3A=3+3^2+3^3+3^4+...+3^{101}\)
\(3A-A=\left(3+3^2+3^3+3^4+...+3^{101}\right)-\left(1+3+3^2+3^3+...+3^{100}\right)\)
\(2A=3^{101}-1\)
\(A=\left(3^{101}-1\right):2\)
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
Bạn tham khảo tại đây nhé: Câu hỏi của Khánh Huyền⁀ᶦᵈᵒᶫ .
Chúc bạn học tốt!
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: \(\Leftrightarrow3^n:27^n=\dfrac{1}{9}\)
\(\Leftrightarrow\left(\dfrac{1}{9}\right)^n=\dfrac{1}{9}\)
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
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
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
2) a) 2x.(22 - 1) = 96 => 2x. 3 = 96 => 2x = 96 : 3 = 32 = 25 => x = 5
b) 7x. (72 + 2. 7-1) = 345 => 7x. \(\frac{345}{7}\) = 345 => 7x = 7 => x = 1
c) 3x-1. (1 + 5) = 162 => 3x-1 . 6 = 162 => 3x-1 = 162 : 6 = 27 = 33 => x - 1 = 3 => x = 3 + 1 = 4
1) a) (33)n = 9.3n => 33n = 32.3n = 32+n => 3n = 2 + n => 3n - n = 2 => 2n = 2 => n = 1
b) 3-2+4+n = 37 => 2 + n = 7 => n = 7 - 2 = 5
c) 2n.(2-1 + 4) = 9.25 => 2n.\(\frac{9}{2}\) = 9.25 => 2n-1 = 25 => n - 1 = 5 => n = 5 + 1 = 6
d) (25)-n.(24)n = 211 => 2-5n. 24n = 211 => 2-5n+4n = 211 => 2-n = 211 => -n = 11 => n = -11