Bài 1:19.C=\(\frac{19^{209}+19}{19^{209}+1}\)=\(\frac{19^{209}+1+18}{19^{209}+1}\)=\(\frac{19^{209}+1}{19^{209}+1}\)+\(\frac{18}{19^{209}+1}\)=1+\(\frac{18}{19^{209}+1}\)19D=\(\frac{19^{210}+19}{19^{210}+1}\)=\(\frac{19^{210}+1+18}{19^{210}+1}\)=\(\frac{19^{210}+1}{19^{210}+1}\)+\(\frac{18}{19^{210}+1}\)=1+\(\frac{18}{19^{210}+1}\).Vì \(\frac{18}{19^{209}+1}\)>\(\frac{18}{19^{210}+1}\)nên 19A>19B\(\Rightarrow\)A>B

19D=\(\frac{\left(19^{209}+1\right).19}{19^{210}+1}=\frac{19^{210}+19}{19^{210}+1}=\frac{\left(19^{210}+1\right)+18}{19^{210}+1}=\frac{19^{210}+1}{19^{210}+1}+\frac{18}{19^{210}+1}=1+\frac{18}{19^{210}+1}\)

Vì 19C>19D nên C>D

Bài 2:Ta có:A=\(\frac{2004}{2005}\)+\(\frac{2005}{2006}\)>\(\frac{2004}{2005+2006}\)+\(\frac{2005}{2005+2006}\)=\(\frac{2004+2005}{2005+2006}\)\(\Rightarrow\)A>B

Có : \(K=\frac{119^{209}+1}{119^{210}+1}<\frac{119^{209}+1+208}{119^{210}+1+208}=\frac{119^{208}.119+119}{119^{209}.119+199}=\frac{119.\left(119^{208}+1\right)}{119.\left(119^{209}+1\right)}=\frac{119^{208}+1}{119^{209}+1}=H\)

=> K < H hay H > K

Chúc bạn học giỏi !!!

Mình chỉ hướng dẫn bạn thôi nhé!

1. Nhân M vs 10 và N vs 10

2.Tách 10M thành 1 + ... và N cũng vậy.

3.So sánh.

Vậy nhé!

CHÚ Ý: bài toán sau: với \(\frac{a}{b}< 1,\)\(\frac{a}{b}< \frac{a+m}{b+m}\)

\(\frac{19^{31}+5}{19^{32}+5}< \frac{19^{31}+5+14}{19^{32}+5+14}=\frac{19^{31}+19}{19^{32}+19}< \frac{19\left(19^{30}+1\right)}{19\left(19^{31}+1\right)}< \frac{19^{30}+1+4}{19^{31}+1+4}=\frac{19^{30}+5}{19^{31}+5}\)

Rõ ràng N<1 nên theo N, nếu \(\frac{a}{b}< 1\Rightarrow\frac{a+n}{b+n}>\frac{a}{b}\)

=> \(\frac{19^{31}+5}{19^{32}+5}< \frac{19^{31}+5.19}{19^{31}+5.19}=\frac{19\left(19^{30}+5\right)}{19\left(19^{31}+5\right)}\)\(=\frac{19^{30}+5}{19^{31}+5}\)

Vậy N<M

Bạn viết rõ bài nhé, đừng tô đen như vậy.

A,Ta có:\(\frac{19}{18}=1+\frac{1}{18};\frac{2011}{2010}=1+\frac{1}{2010}\)

      Vì \(\frac{1}{18}>\frac{1}{2010}\Rightarrow\frac{19}{18}>\frac{2011}{2010}\)

B,ta có:\(1-\frac{72}{73}=\frac{1}{3};1-\frac{98}{99}=\frac{1}{99}\)

       Vì \(\frac{1}{3}>\frac{1}{99}\Rightarrow\frac{72}{73}< \frac{98}{99}\)

C,Vì \(\frac{7}{9}< 1< \frac{19}{17}\Rightarrow\frac{7}{9}< \frac{19}{17}\)

k cho tui rùi tui giải cho nghe !​

tôi ko thik sự giả dối:thôi thì bà làm luôn đi bắt ng khác k làm chi?

ta có : \(\dfrac{1}{2010}>0\) và \(\dfrac{-7}{19}< 0\)

\(\Rightarrow\dfrac{1}{2010}>\dfrac{-7}{19}\)

\(\dfrac{1}{2010}\) > \(\dfrac{-7}{19}\)