\(\dfrac{-592}{591}=-1-\dfrac{1}{591}\)  ; \(\dfrac{-387}{386}=-1-\dfrac{1}{386}\)

Thấy :  \(\dfrac{1}{386}>\dfrac{1}{591}\Rightarrow-1-\dfrac{1}{386}< -1-\dfrac{1}{591}\)  \(\Rightarrow\dfrac{-387}{386}< \dfrac{-592}{591}\)

k) \(\dfrac{-209}{115}\) và \(\dfrac{-241}{131}\)

Ta có : \(\left(-209\right)\cdot131=-27379\)

            \(\left(-241\right)\cdot115=-27715\)

Mà \(\left(-27715\right)< \left(-27379\right)\)

\(\Rightarrow\) \(\dfrac{-209}{115}>\dfrac{-241}{131}\)

\(\dfrac{-251}{138}=\dfrac{-276+25}{138}=-2+\dfrac{25}{138}\)

\(\dfrac{-317}{171}=\dfrac{-342+25}{171}=-2+\dfrac{25}{171}\)

Thấy : \(\dfrac{25}{138}>\dfrac{25}{171}\Rightarrow\dfrac{-251}{138}>\dfrac{-317}{171}\)

-241/131 > -209/115

A>B

A > B

2 ^69 < 5 ^31

ai nhanh nhất mình k cho 

2^100=(2^10)^10=1024^10>1000^10=10^30

2^100=2^31.2^6.2^63=2^31.64.512^7<2^31.5^3.(5^4)^7=2^31.5^31=10^31

10^30<2^100<10^31

Nên 2^100 có 31 chữ số

⇒⇒dpcm

\(\dfrac{7^{45}}{7}=7^{44}\)

\(49.7^{20}=7^2.7^{20}=7^{22}\)

\(\Rightarrow\dfrac{7^{45}}{7}=7^{44}>7^{22}=49.7^{20}\)

7⁴⁵ : 7 ²³ =  7²² 

49 \(\times\) 7²⁰ = 7² \(\times\) 7²⁰ = 7²²

7⁴⁵ : 7²³ = 49 \(\times\) 7²⁰

ai giúp tôi đi ạ , càng nhanh càng tốt ạ !

23,55 : 2,5 = 9,42 
1,3 : 0,25 = 5,2 

\(31^{12}=\left(31^3\right)^4=29791^4\left(1\right)\)

\(27^{20}=\left(27^5\right)^4=14348907^4\left(2\right)\)

từ (1) và (2) => 31^12 <27^20

31\(^{12}\)<32\(^{12}\)=(2\(^5\))\(^{12}\)=2\(^{60}\)   (1)

27\(^{12}\)=(3\(^3\))\(^{20}\)=3\(^{60}\)                  (2)

=> 27\(^{20}\)>31\(^{12}\)