a) C = 27^5 = (3^3)^5 = 3^15

  D = 243 ^3 = (3^5)^3 = 3^15

VẬy C = D 

C=27^4=(3^3)5=3^15

D=243^3=(3^5)^3=3^15

=>C=D

a) \(9=6+3=6+\sqrt{9}\)

\(6+2\sqrt{2}=6+\sqrt{8}\)

\(\sqrt{8}< \sqrt{9}\) nên \(6+\sqrt{8}=6+2\sqrt{2}< 6+\sqrt{9}=9\)

b) \(\left(\sqrt{2}+\sqrt{3}\right)^2=5+2\sqrt{6}=5+\sqrt{24}\)

\(3^2=9=5+4=5+\sqrt{16}\)

\(\sqrt{16}< \sqrt{24}\Rightarrow3^2< \left(\sqrt{2}+\sqrt{3}\right)^2\Rightarrow3< \sqrt{2}+\sqrt{3}\)

c) \(9+4\sqrt{5}=\left(2+\sqrt{5}\right)^2\)

\(16=\left(2+2\right)^2=\left(2+\sqrt{4}\right)^2\)

\(\sqrt{4}< \sqrt{5}\Rightarrow2+\sqrt{4}< 2+\sqrt{5}\Rightarrow\left(2+\sqrt{4}\right)^2=16< \left(2+\sqrt{5}\right)^2=9+4\sqrt{5}\)

d) \(\left(\sqrt{11}-\sqrt{3}\right)^2=14-2\sqrt{33}=14-\sqrt{132}\)

\(2^2=14-10=14-\sqrt{100}\)

\(\sqrt{100}< \sqrt{132}\Leftrightarrow-\sqrt{100}>-\sqrt{132}\Leftrightarrow14-\sqrt{100}>14-\sqrt{132}\)

\(\Rightarrow2>\sqrt{11}-\sqrt{3}\)

a) 2000/ 2002 > 2001/2002

b) 3/5 < 4/9

c) 16/60 < 31/90

cái này mà ko quy đồng thì mới là chuyện lạ có thật!!!

16/27>1529

1995/1996<1996/1997

327/326<326/325

43/6>39/9

nhớ nhé

nhớ tích đúng cho mình nhé đảm bảo 1000000000000000000000000000000000000000000000000000000000000%là đúng

vì sao 16/27lại lơn hơn 15/29

câu c là (2\(^2\))\(^3\)và 2 mũ 2 mũ 3 nha

giúp mk nhanh 1 chút mk cần rất rất gấp

\(a,2^{24}=\left(2^3\right)^8=8^8< 9^8=\left(3^2\right)^8=3^{16}\\ b,16^{11}=\left(2^4\right)^{11}=2^{44}< 2^{45}=\left(2^5\right)^9=32^9\\ \Rightarrow\left(-16\right)^{11}>\left(-32\right)^9\\ c,\left(2^2\right)^3=2^6< 2^8=2^{2^3}\)