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Bài 2:
1: \(5^n+5^{n+2}=650\)
\(\Leftrightarrow5^n\cdot26=650\)
\(\Leftrightarrow5^n=25\)
hay x=2
2: \(32^{-n}\cdot16^n=1024\)
\(\Leftrightarrow\dfrac{1}{32^n}\cdot16^n=1024\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^n=1024\)
hay n=-10
13: \(9\cdot27^n=3^5\)
\(\Leftrightarrow3^{3n}=3^5:3^2=3^3\)
=>3n=3
hay n=1
a) ta có:
\(n^2+1⋮n+1\)
\(\Rightarrow\left(n^2-1\right)+2⋮n+1\)
\(\Rightarrow\left(n-1\right)\left(n+1\right)+2⋮n+1\)
\(\Rightarrow2⋮n+1\)
\(\Rightarrow n+1\in\left\{-1;1;-2;2\right\}\)
\(\Rightarrow x\in\left\{-2;0;-3;1\right\}\)
a)\(\left(\dfrac{1}{2}\right)^n=\dfrac{1}{32}\)
=>\(\left(\dfrac{1}{2}\right)^n=\left(\dfrac{1}{2}\right)^5\)
=>n=5
b)\(\left(\dfrac{343}{125}\right)=\left(\dfrac{7}{5}\right)^n\)
=>\(\left(\dfrac{7}{5}\right)^3=\left(\dfrac{7}{5}\right)^n\)
=>n=3
c)\(\dfrac{16}{2^n}=2\)
=>2n=\(\dfrac{16}{2}\)
=>2n=8
=>2n=23
=>n=3
d)\(\dfrac{\left(-3\right)^n}{81}=-27\)
=>(-3)n=-27.81
=>(-3)n=-2187
=>(-3)n=(-3)7
=>n=7
e)8n:2n=4
=>(23)n:2n=4
=>23n:2n=4
=>23n-n=4
=>22n=4
=>22n=22
=>2n=2
=>n=1
f)32.3n=35
=>3n=35:32
=>3n=35-2
=>3n=33
=>n=3
g) (22:4).2n=4
=>1.2n=22
=>n=2
h)3-2.34.3n=37
=>\(\left(\dfrac{1}{3}\right)^2\).34.3n=37
=>32.3n=37
=>32+n=37
=>2+n=7
=>n=5
a) 9.27n = 35
=> 32.33n = 35
=> 32 + 3n = 35
=> 2 + 3n = 5
=> 3n = 5 - 2
=> 3n = 3
=> n = 1
b) (23 : 4).2n = 4
=> 2.2n = 4
=> 2n = 4 : 2
=> 2n = 2
=> n = 1
c) 3-2.34 . 3n = 37
=> 3-2 + 4 + n = 37
=> 32 + n = 37
=> 2 + n = 7
=> n = 7 - 2 = 5
d) 2-1.2n + 4.2n = 9.25
=> (1/2 + 4).2n = 9.25
=> 9/2.2n = 9.25
=> 2n = 9.25 : 9/2
=> 2n = 26
=> n = 6
\(a,9\cdot27^n=3^5\)
\(\Rightarrow9\cdot27^n=243\)
\(\Rightarrow27^n=243:9=27\)
\(\Rightarrow27^n=27^1\)
\(\Rightarrow x=1\)
\(b,\left(2^3:4\right)\cdot2^n=4\)
\(\Rightarrow\left(8:4\right)\cdot2^n=4\)
\(\Rightarrow2\cdot2^n=4\)
\(\Rightarrow2^n=4:2=2\)
\(\Rightarrow n=1\)
\(c,3^{-2}\cdot3^4\cdot3^n=3^7\)
\(\Rightarrow3^2\cdot3^n=3^7\)
\(\Rightarrow3^n=3^7:3^2=3^5\)
\(\Rightarrow n=5\)
\(d,2^{-1}\cdot2^n+4\cdot2^n=9\cdot2^5\)
\(\Rightarrow2^n\cdot\left(2^{-1}+4\right)=9\cdot32\)
\(\Rightarrow2^n\cdot\frac{9}{2}=288\)
\(\Rightarrow2^n=288:\frac{9}{2}=64\)
\(\Rightarrow2^n=2^6\)
\(\Rightarrow n=6\)
a) \(9.27^n=3^5\Rightarrow3^2.\left(3^3\right)^n=3^5\)
\(\Rightarrow3^2.3^{3n}=3^5\Rightarrow3^{5n}=3^5\)
\(\Rightarrow5n=5\Rightarrow n=1\)
b)\(\left(2^3:4\right).2^n=4\Rightarrow\left(2^3:2^2\right).2^n=2^2\)
\(\Rightarrow2.2^n=2^2\Rightarrow2^{1+n}=2^2\)
\(\Rightarrow1+n=2\Rightarrow n=1\)
c)\(3^2.3^4.3^n=3^7\Rightarrow3^{6+n}=3^7\)
\(\Rightarrow6+n=7\Rightarrow n=1\)
d)\(2^{-1}.2^n+4.2^n=9.2^5\)
\(\Rightarrow2^n\left(2^{-1}+4\right)=3^2.2^5\)
\(\Rightarrow\)\(2^n\left(\frac{1}{2}+4\right)=3^2.2^5\)
\(\Rightarrow\)\(2^n.\frac{3^2}{2}=3^2.2^5\)
\(\Rightarrow\)\(2^{n-1}.3^2=3^2.2^5\)
\(\Rightarrow n-1=5\Rightarrow n=6\)
e)\(243\ge3^n\ge9.3^2\)
\(\Rightarrow3^5\ge3^n\ge3^2.3^2\)
\(\Rightarrow3^5\ge3^n\ge3^4\)
\(\Rightarrow5\ge n\ge4\Rightarrow5;4\)
f)\(2^{n+3}.2^n=128\)
\(\Rightarrow2^{n+3+n}=2^7\)
\(\Rightarrow2^{2n+3}=2^7\)
\(\Rightarrow2n+3=7\Rightarrow2n=4\Rightarrow n=2\)
Hok tối
\(b,3^{n+2}-2^{n+2}+3^n-2^n⋮10\)
Ta có: \(3^{n+2}-2^{n+2}+3^n-2^n\)
\(=3^n.9-2^n.4+3^n-2^n\)
\(=3^n.10-2^n.5\)
Với: \(n\ge1\Rightarrow2^n⋮2\Rightarrow2^n.5⋮10\)
\(3^n.10⋮10\)
\(\Rightarrow3^n.10-2^n.5⋮10\)
\(\Rightarrow\)Ta có đpcm (viết ra cái đề ý)
\(d,7^6+7^5-7^4⋮11\)
Ta có: \(7^6+7^5-7^4=7^4\left(7^2+7-1\right)\)
\(=7^4\left(49+7-1\right)\)
\(=7^4.55\)
Trong tích có thừa số \(55⋮11\)
\(\Rightarrow\)Ta có đpcm (viết ra cái đề ý)
Đề yêu cầu chứng tỏ \(3^{n+2}-2^{n+4}+3^n+2^n⋮30\forall n\) nguyên dương à bạn?
\(3^{n+2}-2^{n+4}+3^n+2^n\)
\(=3^n.9-2^n.16+3^n+2^n\)
\(=\left(3^n.9+3^n\right)+\left(2^n-2^n.16\right)\)
\(=3^n.10-15.2^n\)
Ta có:
\(\left\{{}\begin{matrix}3^n⋮3\\10⋮10\end{matrix}\right.\Rightarrow3^n.10⋮30\) (1)
\(\left\{{}\begin{matrix}15⋮15\\2^n⋮2\end{matrix}\right.\Rightarrow15.2^n⋮30\) (2)
Từ (1) và (2) \(\Rightarrow3^n.10-15.2^n⋮30\)
\(\Rightarrowđpcm.\)
\(1+1=3\)