Ta có:

\(64=2^6=2^{2\cdot3}=\left(2^2\right)^3=4^3\)

⇒ Chọn B 

Ta có:

\(\left(\dfrac{1}{10}\right)^{15}=\left[\left(\dfrac{1}{10}\right)^3\right]^5=\left(\dfrac{1}{1000}\right)^5\)

\(\left(\dfrac{3}{10}\right)^{20}=\left[\left(\dfrac{3}{10}\right)^4\right]^5=\left(\dfrac{81}{10000}\right)^5\)

Ta thấy: \(\dfrac{1}{1000}< \dfrac{81}{10000}\)

\(\Rightarrow\left(\dfrac{1}{1000}\right)^5< \left(\dfrac{81}{10000}\right)^5\)

\(\Rightarrow\left(\dfrac{1}{10}\right)^{15}< \left(\dfrac{3}{10}\right)^{20}\)

A) \(0,75=\dfrac{75}{100}=\dfrac{75:25}{100:25}=\dfrac{3}{4}\)

B) \(0,8=\dfrac{8}{10}=\dfrac{8:2}{10:2}=\dfrac{4}{5}\)

C) \(2,15=\dfrac{215}{100}=\dfrac{215:5}{100:5}=\dfrac{43}{20}\)

Ta có:  \(6,7< y< 6,8\)

3 số thập phân thay vào y được là:

\(y\in\left\{6,71;6,72;6,75\right\}\)

\(x^2=x^3\)

\(\Rightarrow x^2-x^3=0\)

\(\Rightarrow x^2\left(1-x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x^2=0\\1-x=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

\(a^m=a^n\)

\(\Rightarrow m=n\)

Với \(a^m=a^n\) mọi \(m=n\) 

Vậy: \(m=n\in\left\{1;2;3;4;...\right\}\)

Ta có:

\(a=2^{100}=2^{4\cdot25}=\left(2^4\right)^{25}=16^{25}\)

\(b=3^{75}=3^{3\cdot25}=\left(3^3\right)^{25}=27^{25}\)

\(c=5^{50}=5^{2\cdot25}=\left(5^2\right)^{25}=25^{50}\)

Ta thấy: 

\(16< 25< 27\)

\(\Rightarrow16^{25}< 25^{25}< 27^{25}\)

\(\Rightarrow2^{100}< 5^{50}< 3^{75}\)

\(\Rightarrow a< c< b\)

\(1+2+3+...+x=465\) (ĐK: \(x\in N\))

\(\Rightarrow\left(x+1\right)\left[\left(x-1\right):1+1\right]:2=465\)

\(\Rightarrow\left(x+1\right)\left(x-1+1\right):2=465\)

\(\Rightarrow\left(x+1\right)\cdot x:2=465\)

\(\Rightarrow x\left(x+1\right)=465\cdot2\)

\(\Rightarrow x\cdot\left(x+1\right)=930\)

Mà: \(30\cdot31=930\)

Nên: \(x=30\) 

Vậy: x=30

\(1\le3^{n+2}\le729\)

\(\Rightarrow3^0\le3^{n+2}\le3^6\)

\(\Rightarrow0\le n+2\le6\)

\(\Rightarrow0-2\le n\le6-2\)

\(\Rightarrow-2\le n\le4\)

Mà: \(n\in N^+\)

\(\Rightarrow0\le n\le4\)

\(\Rightarrow n\in\left\{0;1;2;3;4\right\}\)

\(5\le5^n\le625\) 

\(\Rightarrow5^1\le5^n\le5^4\)

\(\Rightarrow1\le n\le4\)

Mà: \(n\in N^+\)

\(\Rightarrow n=\left\{1;2;3;4\right\}\)