Bài 1:
a)1/9 x 27ⁿ= 3ⁿ

1/9=3ⁿ:27ⁿ

3ⁿ:27ⁿ=1/9

1ⁿ/9ⁿ=1/9

=>n=1

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

=>2^n-1=2⁵=>n-1=5=>n=5+1=6

vậy......

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8.2ⁿ +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

a) \(k=\frac{2^{11}.9^2}{3^5.16^2}=\frac{2^{11}.\left(3^2\right)^2}{3^5.\left(2^4\right)^2}=\frac{2^{11}.3^4}{3^5.2^8}=\frac{8.1}{3.1}=\frac{8}{3}\)

b)  \(N=\frac{9^3.27^2}{6^2.3^{10}}=\frac{\left(3^2\right)^3.\left(3^3\right)^2}{\left(2.3\right)^2.3^{10}}=\frac{3^6.3^6}{2^2.3^2.3^{10}}=\frac{3^{12}}{4.3^{12}}=\frac{1}{4}\)

Câu 2: 

Ta có: \(x^2=1\)

=>x=1 hoặc x=-1

=>x là số hữu tỉ

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

giúp mình với

thanhks

Bài 2:\(A=\frac{n+1}{n-2009}=\frac{n-2009+2010}{n-2009}=\frac{n-2009}{n-2009}+\frac{2010}{n-2009}=1+\frac{2010}{n-2009}\)

Để A có giá trị lớn nhất \(1+\frac{2010}{n-2009}\)cũng có giá trị lớn nhất =>\(\frac{2010}{n-2009}\)cũng có giá trị lớn nhất => \(n-2009\inƯ\left(2010\right)\)

và \(n-2009\in N\left(n\in Z\right)\)và bé nhất (để\(\frac{2010}{n-2009}\)lớn nhất)

=>n - 2009 = 1 =>n = 2010

Thay n = 2010 vào \(1+\frac{2010}{n-2009}\)ta được: \(1+\frac{2010}{2010-2009}=1+2010=2011\)

Vậy giá trị lớn nhất của A là 2011 khi n=2010

Bài 1:\(A=\frac{5-2n}{n+3}=\frac{9-4+2n}{n+3}=\frac{9}{n+3}-\frac{4+2n}{n+3}=\frac{9}{n+3}-2\)

Để \(A\in N\)thì\(\frac{9}{n+3}-2\in N\Rightarrow\frac{9}{n+3}\in N\Rightarrow n+3\inƯ\left(9\right)\)

Ta có bảng sau:

1)

\(\dfrac{2n+7}{n+1}=\dfrac{2\left(n+1\right)+5}{n+1}=2+\dfrac{5}{n+1}\)

Để \(A\in Z\) thì 5 \(⋮\left(n+1\right)\)

Bảng:

Vậy.....

2)

P = \(\dfrac{-7}{78}.x\)

* Khi P > 0

<=> \(\dfrac{-7}{78}.x\) > 0 => x < 0

* Khi P = 0 <=> x = 0

* Khi P < 0 <=> \(\dfrac{-7}{78}.x\) < 0 =>x > 0

TKS

a) ta có:

\(n^2+1⋮n+1\)

\(\Rightarrow\left(n^2-1\right)+2⋮n+1\)

\(\Rightarrow\left(n-1\right)\left(n+1\right)+2⋮n+1\)

\(\Rightarrow2⋮n+1\)

\(\Rightarrow n+1\in\left\{-1;1;-2;2\right\}\)

\(\Rightarrow x\in\left\{-2;0;-3;1\right\}\)