$\begin{array}{l}5)\\a)\ \text{Ta có :}\\A=\dfrac{10^7+5}{10^7-8}=\dfrac{\left(10^7-8\right)+13}{10^7-8}=1+\dfrac{13}{10^7-8}\\B=\dfrac{10^8+6}{10^8-7}=\dfrac{\left(10^8-7\right)+13}{10^8-7}=1+\dfrac{13}{10^8-7}\\\text{- Ta có : $10^7<10^8$ và $-8<-7$}\\\to 10^7-8<10^8-7\\\to\dfrac{13}{10^7-8}>\dfrac{13}{10^8-7}\\\to 1+\dfrac{13}{10^7-8}>1+\dfrac{13}{10^8-7}\\\to A>B\\\,\\b)\ \text{Ta có :}\\A=\dfrac{10^8+2}{10^8-1}=\dfrac{\left(10^8-1\right)+3}{10^8-1}=1+\dfrac{3}{10^8-1}\\B=\dfrac{10^8}{10^8-3}=\dfrac{\left(10^8-3\right)+3}{10^8-3}=1+\dfrac{3}{10^8-3}\\\text{- Ta có : $10^8-1>10^8-3$}\\\to\dfrac{3}{10^8-1}<\dfrac{3}{10^8-3}\\\to 1+\dfrac{3}{10^8-1}<1+\dfrac{3}{10^8-3}\\\to A<B\\\,\\6)\\a)\ \text{Ta có :}\\A=\dfrac{19^{20}+5}{19^{20}-8}=\dfrac{\left(19^{20}-8\right)+13}{19^{20}-8}=1+\dfrac{13}{19^{20}-8}\\B=\dfrac{19^{21}+6}{19^{21}-7}=\dfrac{\left(19^{21}-7\right)+13}{19^{21}-7}=1+\dfrac{13}{19^{21}-7}\\\text{- Ta có : $19^{20}<19^{21}$ và $-8<-7$}\\\to 19^{20}-8<19^{21}-7\\\to\dfrac{13}{19^{20}-8}>\dfrac{13}{19^{21}-7}\\\to 1+\dfrac{13}{19^{20}-8}>1+\dfrac{13}{19^{21}-7}\\\to A>B\\\,\\b)\ \text{Ta có :}\\\dfrac A{100}=\dfrac{100^{2009}+1}{100.\left(100^{2008}+1\right)}=\dfrac{100^{2009}+1}{100^{2009}+100}=1-\dfrac{99}{100^{2009}+100}\\\dfrac B{100}=\dfrac{100^{2010}+1}{100.\left(100^{2009}+1\right)}=\dfrac{100^{2010}+1}{100^{2010}+100}=1-\dfrac{99}{100^{2010}+100}\\\text{- Vì $2009<2010$}\\\to 100^{2009}<100^{2010}\\\to 100^{2009}+100<100^{2010}+100\\\to\dfrac{99}{100^{2009}+100}>\dfrac{99}{100^{2010}+100}\\\to 1-\dfrac{99}{100^{2009}+100}<1-\dfrac{99}{100^{2010}+100}\\\to\dfrac A{100}<\dfrac B{100}\\\to A<B\end{array}$
