Ta có: \(64< 4^n\le256\)

\(\Leftrightarrow4^3< 4^n\le4^4\)

\(\Rightarrow3< n\le4\)

Mà n là STN => n = 4

Ta có : 64 < 4ⁿ ≤ 256 (với n ∈ N)

=> 4³<4ⁿ ≤ 4⁴

=> 3 < n ≤ 4

=> n = 4 (với n ∈ N)

\(54⋮6\Rightarrow54.67⋮6\)

\(108⋮6\Rightarrow108.25⋮6\)

\(54.67-108.25\Rightarrow A-B⋮6\)

Ta có : 54\(⋮\)6 => 54.67\(⋮\)6

            108\(⋮\)6 => 108.25\(⋮\)6

=> 54.67-108.25\(⋮\)6

Ta có: 6x:(16-2²)=7³:7²

=> 6x:12=7

=> 6x=7.12

=> 6x=84

=> x=84:6

=> x=14

6+x:12=7

6+X=7.12

X=84-6

X=78

\(216\le6^n< 1296\)

\(6^3\le6^n< 6^4\)

\(\Rightarrow n=3\)

\(216=6^3;1296=6^4\)

\(\Rightarrow6^3\le6^n< 6^4\)

\(n=3\)

26 - 3(x + 1) = 14

3(x + 1) = 12

x + 1 = 4

x = 3

26-3(x+1)=14

\(\Rightarrow\)3(x+1)=26-14

\(\Rightarrow\)3(x+1)=12

\(\Rightarrow\)x+1= 12 : 3

\(\Rightarrow\)x+1= 4

\(\Rightarrow\)x = 4-1

\(\Rightarrow\)x=3

Vậy x = 3

\(M=1+\frac{1}{2}+\frac{1}{2^2}+......+\frac{1}{2^{100}}\)

\(\Rightarrow2M=2+1+\frac{1}{2}+..........+\frac{1}{2^{99}}\)

\(\Rightarrow2M-M=M=2-\frac{1}{2^{100}}\)

Ta có M = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\)

=> 2M = \(2+1+\frac{1}{2}+...+\frac{1}{2^{99}}\)

Khi đó 2M - M = \(\left(2+1+\frac{1}{2}+...+\frac{1}{2^{99}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\right)\)

           =>  M \(=2-\frac{1}{2^{100}}\)

mình ko biết

Bài 1: 

a,(2x-15):13+51=64

=> (2x-15):13=64-51

=> (2x-15):13=13

=>(2x-15)=1

=> 2x =16

=> x = 8

Vậy: x= 8

a)\(17\inℤ\left|x-1⋮1\right|\Leftrightarrow x=0\)

b)\(28⋮28;14;7;4;2;1\left|x-1⋮28;14;7;4;2;1\right|\Rightarrow x=29;15;8;5;3;2\)

c)\(2x+5\left|min=1\right|x-2⋮7|\Rightarrow x=9\)

d)\(2x+3\left|\in\left(B\right)2x-1\right|min=1|\Rightarrow x=1\)

theo mình là rỗng