1) tìm số tự nhiên n biết rằng:
a,( 2 n + 1)3 = 27 b, ( n - 2 )2 = ( n - 2 )4
2)Tìm số tự nhiên n biết :
a, 2n.16 = 27 b, ( n - 2 )2 = ( n - 2)4
3)tìm số tự nhiên n, biết
a,3n = 27 b, 5n =625 c, 12n= 144
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1) 274 x 8110
= (33)4 x (34)10
= 33 x 4 x 34 x 10
= 312 x 340
= 352
2) a) n16 = n3
=> n16 - n3 = 0
=> n3 x (n13 - 1) = 0
=> \(\orbr{\begin{cases}n^3=0\\n^{13}-1=0\end{cases}\Rightarrow\orbr{\begin{cases}n^3=0^3\\n^{13}=1^3\end{cases}\Rightarrow}\orbr{\begin{cases}n=0\\n=1\end{cases}}}\)
Vậy n = 0 hoặc n = 1
Cách tui đúng nhất thề luôn
a)2n*16=128
=>2n=128:16
=>2n=8
=>n=4
b)3n*9=27
=>3n=27:9
=>3n=3
=>n=1
c)(2n+1)3=27
=>(2n+1)3=33
=>2n+1=3
=>2n=2
=>n=1
a) 2n.16 = 128
32n = 128
n = 128 : 32
n = 4
Vậy n = 4
b) 3n.9=27
27n = 27
n = 27:27
n = 1
Vậy n = 1
c) (2n + 1)3 = 27
(2n + 1)3 = 33
=> 2n + 1 = 3
=> 2n = 3 - 1 = 2
=> n = 2 : 2 = 1
Vậy n = 1
Bài 6 :
a) \(\dfrac{625}{5^n}=5\Rightarrow\dfrac{5^4}{5^n}=5\Rightarrow5^{4-n}=5^1\Rightarrow4-n=1\Rightarrow n=3\)
b) \(\dfrac{\left(-3\right)^n}{27}=-9\Rightarrow\dfrac{\left(-3\right)^n}{\left(-3\right)^3}=\left(-3\right)^2\Rightarrow\left(-3\right)^{n-3}=\left(-3\right)^2\Rightarrow n-3=2\Rightarrow n=5\)
c) \(3^n.2^n=36\Rightarrow\left(2.3\right)^n=6^2\Rightarrow\left(6\right)^n=6^2\Rightarrow n=6\)
d) \(25^{2n}:5^n=125^2\Rightarrow\left(5^2\right)^{2n}:5^n=\left(5^3\right)^2\Rightarrow5^{4n}:5^n=5^6\Rightarrow\Rightarrow5^{3n}=5^6\Rightarrow3n=6\Rightarrow n=3\)
Bài 7 :
a) \(3^x+3^{x+2}=9^{17}+27^{12}\)
\(\Rightarrow3^x\left(1+3^2\right)=\left(3^2\right)^{17}+\left(3^3\right)^{12}\)
\(\Rightarrow10.3^x=3^{34}+3^{36}\)
\(\Rightarrow10.3^x=3^{34}\left(1+3^2\right)=10.3^{34}\)
\(\Rightarrow3^x=3^{34}\Rightarrow x=34\)
b) \(5^{x+1}-5^x=100.25^{29}\Rightarrow5^x\left(5-1\right)=4.5^2.\left(5^2\right)^{29}\)
\(\Rightarrow4.5^x=4.25^{2.29+2}=4.5^{60}\)
\(\Rightarrow5^x=5^{60}\Rightarrow x=60\)
c) Bài C bạn xem lại đề
d) \(\dfrac{3}{2.4^x}+\dfrac{5}{3.4^{x+2}}=\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{10}}\)
\(\Rightarrow\dfrac{3}{2.4^x}-\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{x+2}}-\dfrac{5}{3.4^{10}}=0\)
\(\Rightarrow\dfrac{3}{2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)+\dfrac{5}{3.4^2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)=0\)
\(\Rightarrow\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)\left(\dfrac{3}{2}+\dfrac{5}{3.4^2}\right)=0\)
\(\Rightarrow\dfrac{1}{4^x}-\dfrac{1}{4^8}=0\)
\(\Rightarrow\dfrac{4^8-4^x}{4^{x+8}}=0\Rightarrow4^8-4^x=0\left(4^{x+8}>0\right)\Rightarrow4^x=4^8\Rightarrow x=8\)
a, 16/2n=2
<=>2n=8
<=>n=4
b, (-3)^n =-27*81=-2187
n=7( vì (-3)^7 =-2187
c, 8^n : 2^n =4
<=> (8:2)^n=4
4^n=4
n=1
\(a,\frac{16}{2^n}=2=>\frac{2^4}{2^n}=2=>2^4:2^n=2=>2^{4-n}=2=>4-n=1=>n=3\)
\(b,\frac{\left(-3\right)^n}{81}=-27=>\frac{\left(-3\right)^n}{3^4}=\left(-3\right)^3=>\frac{\left(-3\right)^n}{\left(-3\right)^4}=\left(-3\right)^3=>\left(-3\right)^{n-4}=\left(-3\right)^3=>n-4=3=>n=7\)
\(c,8^n:2^n=4=>\left(8:2\right)^n=4=>4^n=4=>n=1\)
a)
\(\left(2n+1\right)^3=27\)
\(\left(2n+1\right)^3=3^3\)
\(2n+1=3\)
\(2n=3+1\)
\(2n=4\)
\(n=4\div2\)
\(n=2\)
b)
\(\left(n+2\right)^2=\left(n+2\right)^4\)
\(\left(n+2\right)^4-\left(n+2\right)^2=0\)
\(\left(n+2\right)^2\cdot\left(n+2\right)^2-\left(n+2\right)^2\cdot1=0\)
\(\left(n+2\right)^2\cdot\left[\left(n+2\right)^2-1\right]=0\)
\(\Rightarrow\left(n+2\right)^2=0hoạc\left(n+2\right)^2-1=0\)
\(\left(n+2\right)^2=0\)
\(n+2=0\)
\(n=0+2\)
\(n=2\)
\(\left(n+2\right)^2-1=0\)
\(\left(n+2\right)^2=0+1\)
\(\left(n+2\right)^2=1\)
\(n+2=1\)
\(n=1+2\)
\(n=3\)
Vậy \(n\in\left\{2;3\right\}\)
a,( 2 n + 1)3 = 27
( 2 n + 1)3 = 33
2n+1=3
2n=3-1
2n=2
n=2:2
n=1
a, 3n = 27
3n = 3 mũ 3
b, 5n = 625
5n = 5 mũ 4
c,12n = 144
12n = 12 mũ 2