2³ < 2ⁿ < 2⁶

=> 3 < n < 6

=> n \(\in\){4 ; 5}

25 < 5ⁿ < 625

=> 5² < 5ⁿ < 5⁴

=> 2 < n < 4

=> n = 3

a) \(2^3< 2^n< 2^6\)

\(\Rightarrow3< n< 6\)

\(\Rightarrow n\in\left\{4;5\right\}\)

b) \(25< 5^n< 625\)

\(\Rightarrow5^2< 5^n< 5^4\)

\(\Rightarrow2< n< 4\)

\(\Rightarrow n=3\)

\(S=\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{2019.2020}\)

\(=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+...\left(\frac{1}{2019}-\frac{1}{2020}\right)\)

\(=1-\frac{1}{2020}=\frac{2019}{2020}\)

\(\left|x-7\right|+x-7=0\Leftrightarrow\left|x-7\right|=7-x\)

\(\Leftrightarrow x-7\le0\Leftrightarrow x\le7\)

Vậy \(x\le7\)

MO=10.2=20cm

       \(\left(5^3+5^4+125^2\right):5^3\)

\(=\left[5^3+5^4+\left(5^3\right)^2\right]:5^3\)

\(=\left(5^3+5^4+5^6\right):5^3\)

\(=5^3\times\left(1+5+5^3\right):5^3\)

\(=1+5+5^3\)

\(=131\)

\(\left(5^3+5^4+125^2\right):5^3=\left(5^3+5^4+5^6\right):5^3\)

\(=5^3\left(1+5+5^3\right):5^3=1+5+125\)

\(=131\)

\(B=12+30+x\)

\(=42+x\)

Để B không chia hết cho 3

\(\Rightarrow42+x\)không chia hết cho 3

Mà 42 chia hết cho 3

\(\Rightarrow x\)sẽ không chia hết cho3

\(\Rightarrow x\)sẽ có dạng : 3K+1 ( k \(\in Z\)) 

                                 3k+2