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

\(N=\dfrac{\dfrac{2.2}{3}\left(\dfrac{1}{5}-\dfrac{1}{7}-\dfrac{1}{11}\right)}{\dfrac{6.6}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}-\dfrac{1}{11}\right)}=\dfrac{4}{\dfrac{3}{\dfrac{36}{2}}}=24\)

\(M=\dfrac{3\cdot\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{6\cdot\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}=\dfrac{3}{6}=\dfrac{1}{2}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{10}{39}=\dfrac{5}{39}\)

\(\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}=\dfrac{5}{39}\)

\(=0,5.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\right)\)

\(=0,5.\left(\dfrac{1}{3}-\dfrac{1}{13}\right)=0,5\cdot\dfrac{10}{39}=\dfrac{5}{39}\)

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

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

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

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\(^{12}\)

Có 2^285 = (2^3)^95 = 8^95

     3^190 = (3^2)^95 = 9^95

Vì 8^95 < 9^95 nên 2^285 < 3^190

2²⁸⁵> 3¹⁹⁰

So sánh :

\(2^{285}\)và \(3^{190}\)

\(\Rightarrow2^{285}>3^{190}\)