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TL:
a.\(2^6.2^n=2^{11}\)
\(2^{6+n}=2^{11}\)
\(\Rightarrow n=5\)
b. \(3^7:3^n=3^4\)
\(3^{7-n}=3^4\)
\(\Rightarrow n=3\)
c.\(2^n.32=2^{10}\)
\(2^{n+5}=2^{10}\)
\(\Rightarrow n=5\)
32 . 3n = 35
=> 2 + n = 5
=> n = 5 - 2
=> n = 3
( 22 : 4 ) . 2n = 4
( 4 : 4 ) . 2n = 22
1 . 2n = 22
=> n = 2
Các câu sau tự làm nhé
a) \(4^n=4096\Rightarrow4^n=4^6\Rightarrow n=6\)
b) \(5^n=15625\Rightarrow5^n=5^6\Rightarrow n=6\)
c) \(6^{n+3}=216\Rightarrow6^{n+3}=6^3\Rightarrow n+3=3\Rightarrow n=0\)
d) \(x^2=x^3\Rightarrow x^3-x^2=0\Rightarrow x^2\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
e) \(3^{x-1}=27\Rightarrow3^{x-1}=3^3\Rightarrow x-1=3\Rightarrow x=4\)
f) \(3^{x+1}=9\Rightarrow3^{x+1}=3^2\Rightarrow x+1=2\Rightarrow x=1\)
g) \(6^{x+1}=36\Rightarrow6^{x+1}=6^2\Rightarrow x+1=2\Rightarrow x=1\)
h) \(3^{2x+1}=27\Rightarrow3^{2x+1}=3^3\Rightarrow2x+1=3\Rightarrow2x=2\Rightarrow x=1\)
i) \(x^{50}=x\Rightarrow x^{50}-x=0\Rightarrow x\left(x^{49}-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^{49}-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x^{49}=1=1^{49}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
4n = 4096
4n = 212
n = 12
5n = 15625
5n = 56
n = 6
6n+3 = 216
6n+3 = 23.33
6n+3 = 63
n + 3 = 3
Bài 1: a) \(M=1+5+5^2+...+5^{100}\)
\(5M=5+5^2+5^3+...+5^{101}\)
\(5M-M=\left(5+5^2+5^3+...+5^{101}\right)-\left(1+5+5^2+...+5^{100}\right)\)
\(4M=5^{101}-1\)
\(M=\frac{5^{101}-1}{4}\)
b) \(N=2+2^2+...+2^{100}\)
\(2N=2^2+2^3+...+2^{101}\)
\(2N-N=\left(2^2+2^3+...+2^{101}\right)-\left(2+2^2+...+2^{100}\right)\)
\(N=2^{101}-2\)
Bài 2:
a) \(16^{32}=\left(2^4\right)^{32}=2^{128}\)
\(32^{16}=\left(2^5\right)^{16}=2^{80}\)
Vì \(2^{128}>2^{80}\Rightarrow16^{32}>32^{16}\)
Bài 1:
a,\(A=3+3^2+3^3+...+3^{2010}\)
\(=\left(3+3^2+3^3+3^4\right)+....+\left(3^{2007}+3^{2008}+3^{2009}+3^{2010}\right)\)
\(=3\left(1+3+3^2+3^3\right)+....+3^{2007}\left(1+3+3^2+3^3\right)\)
\(=3.40+...+3^{2007}.40\)
\(=40\left(3+3^5+...+3^{2007}\right)⋮40\)
Vì A chia hết cho 40 nên chữ số tận cùng của A là 0
b,\(A=3+3^2+3^3+...+3^{2010}\)
\(3A=3^2+3^3+...+3^{2011}\)
\(3A-A=\left(3^2+3^3+...+3^{2011}\right)-\left(3+3^2+3^3+...+3^{2010}\right)\)
\(2A=3^{2011}-3\)
\(2A+3=3^{2011}\)
Vậy 2A+3 là 1 lũy thừa của 3
câu 1:
(24 .2 .32 .23) : ( 22 . 32 . 4 )
=( 24 . 2. 25 . 23) : ( 22 . 25 . 22 )
= 213 . 29
= 222
câu 2 tương tự nha bạn
