Tìm số tự nhiên n , biết
a) \(\dfrac{2^n}{32}\) = 4
b) \(^{^{ }}27^n\) . 9\(^n\) = 9\(^{27}\) : 81
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2^n/32 = 4 => 2^n = 4 . 32 = 128 => n =7
27^n . 9^n = 9^27 . 81
=> (27.9)^n = 9^27 . 9^2
=> 243^n = 9^54
=> 243^n = 243^1458
vay n=1458
1/9 . 3^4 . 3^n+1 = 9^4
=> 9 . 3^n+1 = 6561
=> 3^n+1 = 6561 /9
=> 3^n+1 = 729
=> n = 5
mk ghi lại đề nha:
27n : 9n = 927 : 81
(27 : 9)n = 927 : 92
\(\Rightarrow\) 3n = 925
\(\Rightarrow\) 3n = (32)25
\(\Rightarrow\) 3n = 350
Vậy n = 50
\(27^n.9^n=9^{27}:81\Rightarrow3^{3n}:3^{2n}=3^{54}:3^4=3^{50}\)
\(\Rightarrow3^{5n}=3^{50}\Rightarrow5n=50\Rightarrow n=\frac{50}{5}=10\)
\(27^n.9^n=9^{27}:81\)
\(3^{3n}.3^{2n}=3^{54}:3^4\)
\(3^{5n}=3^{50}\)
=> 5n = 50
=> n = 10
2, 100^2+200^2+300^2+..+1000^2
=100^2+2^2×100^2+3^2×100^2+...+100^2×10^2
=100^2×( 1^2+2^2+3^2+..+10^2)
=100^2×385
= 3850000
a)
b,
\(\dfrac{\left(-3\right)^n}{81}=-27\Rightarrow\dfrac{\left(-3\right)^n}{\left(-3\right)^4}=-27\Rightarrow\left(-3\right)^{n-4}=\left(-3\right)^3\Rightarrow n-4=3\Rightarrow n=7\)
c,\(8^n:2^n=4\Rightarrow4^n=4\Rightarrow n=1\)
=> (-3)n-4 = (-3)3
=> n - 4 = 3 => n = 7
c) 8n : 2n = 4
4n = 4.
\(a,\frac{16}{2^n}=2=>\frac{2^4}{2^n}=2=>2^4:2^n=2=>2^{4-n}=2=>4-n=1=>n=3\)
\(b,\frac{\left(-3\right)^n}{81}=-27=>\frac{\left(-3\right)^n}{3^4}=\left(-3\right)^3=>\frac{\left(-3\right)^n}{\left(-3\right)^4}=\left(-3\right)^3=>\left(-3\right)^{n-4}=\left(-3\right)^3=>n-4=3=>n=7\)
\(c,8^n:2^n=4=>\left(8:2\right)^n=4=>4^n=4=>n=1\)
a,
\(\dfrac{2^n}{32}=4\\ 2^n:2^5=2^2\\ 2^n=2^2\cdot2^5\\ 2^n=2^7\\ n=7\)
b,
\(27^n\cdot9^n=9^{27}:81\\ \left(3^3\right)^n\cdot\left(3^2\right)^n=\left(3^2\right)^{27}:3^4\\ 3^{3n}\cdot3^{2n}=3^{54}:3^4\\ 3^{3n+2n}=3^{50}\\ 3^{5n}=3^{50}\\5n=50\\ n=10 \)
a/ \(\dfrac{2^n}{32}=4\)
\(\Leftrightarrow\dfrac{2^n}{2^5}=2^2\)
\(\Leftrightarrow2^n=2^7\)
\(\Leftrightarrow n=7\)
Vậy ...
b/ \(27^n.9^n=9^{27}:81\)
\(\Leftrightarrow3^{3n}.3^{2n}=3^{54}:3^4\)
\(\Leftrightarrow3^{2n+3n}=3^{50}\)
\(\Leftrightarrow2n+3n=50\)
\(\Leftrightarrow5n=50\)
\(\Leftrightarrow n=10\)
Vậy ...
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