Đặt: \(A=9^9+9^{10}+...+9^{20}\) 

\(9A=9\cdot\left(9^9+9^{10}+...+9^{20}\right)\)

\(9A=9^{10}+9^{11}+...+9^{21}\)

\(9A-A=9^{10}+9^{11}+...+9^{21}-9^9-9^{10}-...-9^{20}\)

\(8A=9^{21}-9^9\)

\(A=\dfrac{9^{21}-9^9}{8}\)

Đặt $A=9^9+9^{10}+...+9^{20}$

$9\cdot A=9\cdot(9^9+9^{10}+...+9^{20})$

$9A=9^{10}+9^{11}+...+9^{21}$

$9A-A=(9^{10}+9^{11}+...+9^{21})-(9^9+9^{10}+...+9^{20})$

$8A=9^{21}-9^9$

$\Rightarrow A=\dfrac{9^{21}-9^9}{8}$

\(4\cdot5^2-3\cdot2^3+3^3\cdot3^2\)

\(=4\cdot25-3\cdot8+3^{2+3}\)

\(=100-24+3^5\)

\(=100-24+243\)

\(=343-24\)

\(=319\)

3n + 5 chia hết cho 2n + 1 

⇒ 2(3n + 5) chia hết cho 2n + 1

⇒ 6n + 10 chia hết cho 2n + 1

⇒ 6n + 3 + 7 chia hết cho 2n + 1

⇒ 3(2n + 1) + 7 chia hết cho 2n + 1

⇒ 7 chia hết cho 2n + 1

⇒ 2n + 1 ∈ Ư(7) = {1; -1; 7; -7} 

⇒ n ∈ {0; -1; 3; -4}

Mà: n ∈ N

⇒ n ∈ {0; 3} 

\(A=24\cdot6^3\)

\(A=4\cdot6\cdot\left(2\cdot3\right)^3\)

\(A=2^2\cdot2\cdot3\cdot2^3\cdot3^3\)

\(A=\left(2^2\cdot2\cdot2^3\right)\cdot\left(3\cdot3^3\right)\)

\(A=2^6\cdot3^4\)

A = 24.6³

A  = 2³.3.2³.3³

A = 2⁶.3⁴

\(A=2+2^2+...+2^{60}\)

\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\)

\(A=2\cdot3+2^3\cdot3+...+2^{59}\cdot3\)

\(A=3\cdot\left(2+2^3+...+2^{59}\right)\)

Vậy A chia hết cho 3 

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\(A=2+2^2+2^3+...+2^{60}\)

\(A=\left(2+2^3\right)+\left(2^2+2^4\right)+...+\left(2^{58}+2^{60}\right)\)

\(A=2\cdot5+2^2\cdot5+...+2^{58}\cdot5\)

\(A=5\cdot\left(2+2^2+...+2^{58}\right)\)

Vậy A ⋮ 5

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\(A=2+2^2+2^3+...+2^{60}\)

\(A=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)

\(A=2\cdot7+2^4\cdot7+...+2^{58}\cdot7\)

\(A=7\cdot\left(2+2^4+...+2^{58}\right)\)

Vậy A ⋮ 7 

\(24\cdot\left(x-16\right)=144\)

\(\Rightarrow x-16=\dfrac{114}{24}\)

\(\Rightarrow x-16=6\)

\(\Rightarrow x=6+16\)

\(\Rightarrow x=22\)

Vậy: \(x=22\)

24 . ( x - 16 ) = 144

        ( x - 16 ) = 144 : 24

         x - 16     = 6 

         x.           =  6 + 16

         x.           = 22

\(10^4\cdot\left[4\cdot\left(5^2-5\right)\right]+25\)

\(=1000\cdot\left[4\cdot\left(25-5\right)\right]+25\)

\(=1000\cdot\left(4\cdot20\right)+25\)

\(=1000\cdot80+25\)

\(=80000+25\)

\(=80025\)

\(\left(\dfrac{1}{2}\right)^3\cdot\left(\dfrac{1}{4}\right)^2\)

\(=\left(\dfrac{1}{2}\right)^3\cdot\left[\left(\dfrac{1}{2}\right)^2\right]^2\)

\(=\left(\dfrac{1}{2}\right)^3\cdot\left(\dfrac{1}{2}\right)^{2\cdot2}\)

\(=\left(\dfrac{1}{2}\right)^3\cdot\left(\dfrac{1}{2}\right)^4\)

\(=\left(\dfrac{1}{2}\right)^{3+4}\)

\(=\left(\dfrac{1}{2}\right)^7\)

Ta có: \(A=1+2+2^2+...+2^{2015}\)

\(2A=2\cdot\left(1+2+2^2+...+2^{2015}\right)\)

\(2A=2+2^2+2^3+...+2^{2016}\)

\(2A-A=2+2^2+...+2^{2016}-1-2-2^2-...-2^{2015}\)

\(A=2^{2016}-1\)

A không thể biết dưới dạng lũy thừa của 8 được 

A=22016−1