\(3^{n+2}\) + \(3^{n+1}\)\(3^n\)
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8 tháng 6 2023

\(3^{n+2}+3^{n+1}-3^n=891\)

\(3^n\times3^2+3^n\times3-3^n=891\)

\(3^n\times\left(9+3-1\right)=891\)

\(3^n\times11=891\)

\(3^n=891\div11\)

\(3^n=81\)

\(3^n=3^4\)

\(n=4\)

8 tháng 6 2023

\(3^{n+2}+3^{n+1}-3^n=891\)

\(\Leftrightarrow3^n.3^2+3^n.3-3^n=891\)

\(\Leftrightarrow3^n\left(3^2+3-1\right)=891\)

\(\Leftrightarrow3^n.11=891\)

\(\Leftrightarrow3^n=81\)

\(\Rightarrow n=4\)

22 tháng 9 2019

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

22 tháng 9 2019

\(a,9\cdot27^n=3^5\)

\(\Rightarrow9\cdot27^n=243\)

\(\Rightarrow27^n=243:9=27\)

\(\Rightarrow27^n=27^1\)

\(\Rightarrow x=1\)

\(b,\left(2^3:4\right)\cdot2^n=4\)

\(\Rightarrow\left(8:4\right)\cdot2^n=4\)

\(\Rightarrow2\cdot2^n=4\)

\(\Rightarrow2^n=4:2=2\)

\(\Rightarrow n=1\)

\(c,3^{-2}\cdot3^4\cdot3^n=3^7\)

\(\Rightarrow3^2\cdot3^n=3^7\)

\(\Rightarrow3^n=3^7:3^2=3^5\)

\(\Rightarrow n=5\)

\(d,2^{-1}\cdot2^n+4\cdot2^n=9\cdot2^5\)

\(\Rightarrow2^n\cdot\left(2^{-1}+4\right)=9\cdot32\)

\(\Rightarrow2^n\cdot\frac{9}{2}=288\)

\(\Rightarrow2^n=288:\frac{9}{2}=64\)

\(\Rightarrow2^n=2^6\)

\(\Rightarrow n=6\)

10 tháng 4 2018

a) \(10^{n+1}-6.10^n\)

\(=10^n.10-6.19^n\)

\(=10^n.\left(10-6\right)\)

\(=10^n.4\)

b) \(2^{n+3}+2^{n+2}-2^{n+1}+2^n\)

\(=2^n.2^3+2^n.2^2-2^n.2+2^n.1\)

\(=2^n.\left(2^3+2^2-2+1\right)\)

\(=2^n.11\)

c) \(90.10^k-10^{k+2}+10^{k+1}\)

\(=90.10^k-10^k.10^2+10^k.10\)

\(=10^k.\left(90-10^2+10\right)\)

\(=0\)

d) \(2,5.5^{n-3}.10+5^n-6.5^{n-1}\)

\(=\dfrac{2,5.5^n.10}{5^3}+5^n-\dfrac{6.5^n}{5}\)

\(=\dfrac{5^n}{5}+5^n-\dfrac{6.5^n}{5}\)

\(=\dfrac{5^n+5^{n+1}-6.5^n}{5}=\dfrac{5^n+5^n.5-6.5^n}{5}=\dfrac{5^n\left(1+5-6\right)}{5}=\dfrac{0}{5}=0\)

24 tháng 7 2017

\(A=\left(-1\right)^{2n}.\left(-1\right)^n.\left(-1\right)^{n+1}\)

\(A=\left(-1\right)^{2n+n+n+1}\)

\(A=\left(-1\right)^{4n+1}\)

\(B=\left(10000-1^2\right).\left(10000-2^2\right)...\left(10000-1000^2\right)\)

\(B=\left(10000-1^2\right)\left(10000-2^2\right)...\left(10000-100^2\right)...\left(10000-1000^2\right)\)

\(B=\left(10000-1^2\right)\left(10000-2^2\right)...\left(10000-10000\right)...\left(10000-1000^2\right)\)

\(B=\left(10000-1^2\right)\left(10000-2^2\right)...0\left(10000-1000^2\right)\)

\(B=0\)

\(C=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)...\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)

\(C=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)...\left(\dfrac{1}{125}-\dfrac{1}{5^3}\right)...\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)

\(C=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)...0....\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)

\(C=0\)

\(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-10^3\right)}\)

\(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-1000\right)}\)

\(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)...0}\)

\(D=1999^0\)

\(D=1\)

26 tháng 5 2017

Bài 1:
a)1/9 x 27n= 3n

1/9=3n:27n

3n:27n=1/9

1n/9n=1/9

=>n=1

26 tháng 5 2017

\(\frac{1}{2}.2^n+4.2^n=9.2^5\Rightarrow2^n\left(\frac{1}{2}+4\right)=288\Rightarrow2^n.\frac{9}{2}=288\Rightarrow2^{n-2}.9=288\Rightarrow2^{n-2}=32\)(dấu "=>" số 3 bn sửa thành 2n-1.9=288=>2n-1=32 nha)

=>2n-1=25=>n-1=5=>n=5+1=6

vậy......

~~~~~~~~~~~~~~~

3 tháng 10 2019

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

25 tháng 7 2017

Bài 3: Tìm x:

a. \(\left(2x-1\right)^4=81\)

\(\Rightarrow\left(2x-1\right)^4=3^4\)

=> 2x - 1 = 3

=> 2x = 4

=> x = 2

b. \(\left(x-2\right)^2=1\)

\(\Rightarrow\) \(\left(x-2\right)^2=1^2\)

=> x - 2 = 1

=> x = 3

c. \(x^{2000}=x\)

=> x = 1

d. \(\left(4x-3\right)^3=-125\)

\(\Rightarrow\left(4x-3\right)^3=\left(-5\right)^3\)

=> 4x - 3 = -5

=> 4x = -2

=> x = \(\dfrac{-1}{2}\)

26 tháng 7 2017

came ơn bạn nhìu!!!!!!!!

24 tháng 8 2017

Tính kiểu lớp 7 hay kiểu lớp 8 v Bo?

24 tháng 8 2017

Vậy Bo dùng máy tính tính đi,dễ mà,máy tính tính đc

12 tháng 12 2017

Ta có :

3n+2 + 3n + 2n+2 + 2n

= 3n . 32 + 3n + 2n . 22 + 2n

= 3n . ( 32 + 1 ) + 2n . ( 22 + 1 )

= 3n . 10 + 2n . 5

= 3n . 10 + 2n-1 . 10

= 10 . ( 3n + 2n-1 ) \(⋮\)10

17 tháng 2 2019

Ta có :3n + 2 + 3 n + 2 n + 2 + 2 n

= 3n . 32 + 3n + 2n . 22 + 2n

= 3n . ( 32 + 1 ) + 2n . ( 22 + 1 )

= 3n . 10 + 2n . 5

= 3n . 10 + 2n-1 . 10

= 10 . ( 3n + 2n-1 ) \(⋮\)10 ( vì \(10⋮10\)\(\left[đpcm\right]\)

13 tháng 3 2017

8.2+2n+1 

=2n .(8+2)

=2n.10 chia hết cho 10

=> 8.2n +2n+1 chia hết cho 10

\(3^{n+3^{ }}-2.3^n+2^{n+5}-7.2^n\)

\(=3^n.\left(3^3-2\right)+2^n\left(2^5-7\right)\)

\(=3^n.25+2^n.25\)

=\(25.\left(3^n+2^n\right)\)chia hết cho 25

=>\(3^{n+3}-2.3^n+2^{n+5}-7.2^n\)

k cho mình nhé

3 tháng 10 2017

Giúp mình nha!

Mai mình nộp rồi.