Lời giải:
$199^{20}< 200^{20}=100^{20}.2^{20}=100^{20}.(2^5)^4=100^{20}.32^4$

$< 100^{20}.100^4=100^{24}$

\(3^{151}< 3^{150}\)

Ta có:

\(3^{150}=\left(3^2\right)^{75}=9^{75}\)

\(2^{225}=\left(2^3\right)^{75}=8^{75}\)

Mà: \(9>8\)

\(\Rightarrow9^{75}>8^{75}\)

\(\Rightarrow3^{150}>2^{225}\)

\(\Rightarrow3^{151}>2^{225}\)

Ta có:

\(15^{12}=\left(3\cdot5\right)^{12}=3^{12}\cdot5^{12}\)

\(81^3\cdot125^5=\left(3^4\right)^3\cdot\left(5^3\right)^5=3^{12}\cdot5^{15}\)

Mà: \(15>12\)

\(\Rightarrow5^{15}>5^{12}\)

\(\Rightarrow3^{12}\cdot5^{15}>3^{12}\cdot5^{12}\)

\(\Rightarrow81^3\cdot125^5>15^{12}\)

Ta có:

\(2^{16}=2^3\cdot2^{13}=8\cdot2^{13}\)

Mà:

\(7< 8\)

\(\Rightarrow7\cdot2^{13}< 8\cdot2^{13}\)

\(\Rightarrow7\cdot2^{13}< 2^{16}\)

Ta có:

\(5^{23}=5\cdot5^{22}\)

Mà:

\(6>5\)

\(\Rightarrow6\cdot5^{22}>5\cdot5^{22}\)

\(\Rightarrow6\cdot5^{22}>5^{23}\)

Ta có:

\(3>2\)

\(\Rightarrow3^{3n}>2^{3n}\) (do n ∈ N) 

Vậy: ... 

3²ⁿ và 2³ⁿ ( n E N*)

Ta có: 3²ⁿ = (3²)ⁿ = 9ⁿ

          2³ⁿ =  (2³)ⁿ = 8ⁿ

Vì 9>8 => 9ⁿ > 8ⁿ hay 3²ⁿ > 2³ⁿ

Vậy 3²ⁿ > 2³ⁿ   

Lời giải:

$4^{30}=(4^3)^{10}=64^{10}> 48^{10}=(2.24)^{10}=2^{10}.24^{10}> 3.24^{10}$

Ta có:

\(3^{210}=\left(3^3\right)^{70}=27^{70}\)

\(2^{350}=\left(2^5\right)^{70}=32^{70}\)

Mà: \(32>27\)

\(\Rightarrow32^{70}>27^{70}\)

\(\Rightarrow2^{350}>3^{210}\)

Ta có: 

\(32^{60}=\left(2^5\right)^{60}=2^{300}=\left(2^3\right)^{100}=8^{100}\)

\(81^{50}=\left(9^2\right)^{50}=9^{100}\)

Mà: \(8< 9\)

\(\Rightarrow8^{100}< 9^{100}\)

\(\Rightarrow32^{60}< 81^{50}\)

Bài 1: 

144 = 2⁴.3²

216 = 2³.3³

160 = 2⁵.5

ƯCLN(144; 216; 160) = 2³ = 8

BCNN(144; 216; 160) = 2⁵.3³.5 = 4320

Bài 2: Gọi số tự nhiên thỏa mãn đề bài là \(x\)

Vì \(x\) \(⋮\) 4 dư 1 nên \(x\) có dạng: \(x\) = 4k + 1 (k \(\in\) N; k_min)

Theo bài ra ta có:

 \(\left\{{}\begin{matrix}4k+1-6⋮9\\4k+1-8⋮11\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}4k-5⋮9\\4k-7⋮11\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}11.\left(4k-5\right)⋮9.11\\9.\left(4k-7\right)⋮11.9\end{matrix}\right.\)

⇒\(\left\{{}\begin{matrix}44k-55⋮99\\36k-63⋮99\end{matrix}\right.\) \(\Rightarrow\)  44k - 55 - (36k - 63)⋮ 99 

⇒ 44k - 55 - 36k + 63 ⋮ 99⇒ 8k + 8 ⋮ 99 ⇒ 8.(k + 1)⋮ 99

⇒ k + 1 ⋮ 99 ⇒ k + 1 \(\in\) {0; 99; 198;...}

   ⇒ k \(\in\) {-1; 98; 197;..;}

Vì k \(\in\) N và k_min ⇒ k = 98

Vậy \(x\) = 4.98 + 1 = 393