so sánh 10^2019+1/10^2018+1 và 10^2018+1/10^2017+1

0 bình luận về “so sánh 10^2019+1/10^2018+1 và 10^2018+1/10^2017+1”

1. Đáp án: (10^2019 + 1)/(10^2018 + 1) > (10^2018 + 1)/(10^2017 + 1)

    

   Giải thích các bước giải:

   Với mọi a > 0 ta có : 100a² + 101a + 1 > 100a² + 20a + 1

   ⇔ 100a(a + 1) + (a + 1) > 100a² + 20a + 1

   ⇔ (100a + 1)(a + 1) > (10a + 1)²

   ⇔ (100a + 1)/(10a + 1) > (10a + 1)/(a + 1) (*)

   Thay a = 10^2017 ⇒ 100a = 10^2019; 10a = 10^2017 vào (*) ta có:

   (10^2019 + 1)/(10^2018 + 1) > (10^2018 + 1)/(10^2017 + 1)

    

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2. Đáp án:

   \[\frac{{{{10}^{2019}} + 1}}{{{{10}^{2018}} + 1}} > \frac{{{{10}^{2018}} + 1}}{{{{10}^{2017}} + 1}}\]

   Giải thích các bước giải:

    Ta có:

   \(\begin{array}{l}
   A = \frac{{{{10}^{2019}} + 1}}{{{{10}^{2018}} + 1}} \Rightarrow \frac{A}{{10}} = \frac{{{{10}^{2019}} + 1}}{{{{10}^{2019}} + 10}} = \frac{{\left( {{{10}^{2019}} + 10} \right) – 9}}{{{{10}^{2019}} + 10}} = 1 – \frac{9}{{{{10}^{2019}} + 10}}\\
   B = \frac{{{{10}^{2018}} + 1}}{{{{10}^{2017}} + 1}} \Rightarrow \frac{B}{{10}} = \frac{{{{10}^{2018}} + 1}}{{{{10}^{2018}} + 10}} = \frac{{\left( {{{10}^{2018}} + 10} \right) – 9}}{{{{10}^{2018}} + 10}} = 1 – \frac{9}{{{{10}^{2018}} + 10}}\\
   {10^{2019}} > {10^{2018}} \Rightarrow {10^{2019}} + 10 > {10^{2018}} + 10 \Rightarrow \frac{9}{{{{10}^{2019}} + 10}} < \frac{9}{{{{10}^{2018}} + 10}} \Leftrightarrow \frac{{ – 9}}{{{{10}^{2019}} + 10}} > \frac{{ – 9}}{{{{10}^{2018}} + 10}}\\
    \Rightarrow \frac{A}{{10}} > \frac{B}{{10}}\\
    \Leftrightarrow A > B \Leftrightarrow \frac{{{{10}^{2019}} + 1}}{{{{10}^{2018}} + 1}} > \frac{{{{10}^{2018}} + 1}}{{{{10}^{2017}} + 1}}
   \end{array}\)

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