2⁹¹ và 5³⁶

 2⁹¹=  (2⁵)¹⁸.2 = 32¹⁸.2 

 5³⁶ = (5²)¹⁸    = 25¹⁸

 32 > 25 ⇒ 32¹⁸ > 25¹⁸ ⇒ 2⁹¹ > 5³⁶

\(2^{91}>2^{90}\)

Ta có:

\(2^{90}=\left(2^5\right)^{18}=32^{18}\)

\(5^{36}=\left(5^2\right)^{18}=25^{18}\)

Mà: \(32>25\)

\(\Rightarrow32^{18}>25^{18}\)

\(\Rightarrow2^{90}>5^{36}\)

\(\Rightarrow2^{91}>5^{36}\)

Ta có:

\(2^{91}>2^{90}=\left(2^5\right)^{18}=32^{18}>25^{18}=\left(5^2\right)^{18}=5^{2.18}=5^{36}\)

Vậy \(2^{91}>5^{36}\).

\(x\)(y - 3) = -12

⇒ \(x\).(3-y) = 12

12 = 2².3

Ư(12) = {-12; -6; -3; -4; -2; -1; 1; 2; 3; 4;6;12}

Lập bảng ta có:

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:

\(a=40=2^3\cdot5\)

\(b=75=5^2\cdot3\)

\(c=105=5\cdot3\cdot7\) 

\(\RightarrowƯCLN\left(a,b,c\right)=5\) 

\(\Rightarrow BCNN\left(a,b,c\right)=5^2\cdot2^3\cdot3\cdot7=4200\)

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