Câu 1: So sánh a/ A = $ \frac{$ 10^{1990}$ +1}{$ 10^{1991}$ +1}$ và B = $ \frac{$ 10^{1991}$ +1}{$ 10^{1992}$ +1}$ b C = $ \frac{2019^{20}$ +1}{2

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1. `Câu` `1:`

   `a)“Ta` `có:`

   `B<1=>B<{10^1991+1+9}/{10^1992+1+9}={10^1991+10}/{10^1992+10}={10(10^1990+1)}/{10(10^1991+1)}={10^1990+1}/{10^1991+1}=A`
                       ` Vậy` `B<A`  

   `b)“Ta` ` có:`

   `C>1=>C>{2019^20+1+2018}/{2019^19+1+2018}={2019^20+2019}/{2019^19+2019}={2019(2019^19+1)}/{2019(2019^18+1)}={2019^19+1}/{2019^18+1}=D`
   ` Vậy ` `C>D`

     

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2. Giải thích các bước giải:

   a. Ta có:

   `10A=(10^1991+10)/(10^1991+1)=1+9/(10^1991+1)`

   `10B=(10^1992+10)/(10^1992+1)=1+9/(10^1992+1)`

   Vì `1+9/(10^1991+1)>1+9/(10^1992+1)` nên `10A>10B⇔A>B`

   Vậy `A>B`

   b. Ta có:

   `1/2019C=(2019^20+1)/(2019^20+2019)=1-2018/(2019^20+2019)`

   `1/2019D=(2019^19+1)/(2019^19+2019)=1-2018/(2019^19+2019)`

   Vì `2018/(2019^20+2019)<2018/(2019^19+2019)`

   nên `1-2018/(2019^20+2019)>1-2018/(2019^19+2019)`

   `⇒1/2019C>1/2019D⇔C>D`

   Vậy `C>D`

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