so sánh 10^2019+1/10^2018+1 và 10^2018+1/10^2017+1 30/07/2021 Bởi Adalynn so sánh 10^2019+1/10^2018+1 và 10^2018+1/10^2017+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) Bình luận
Đáp án: \[\frac{{{{10}^{2019}} + 1}}{{{{10}^{2018}} + 1}} > \frac{{{{10}^{2018}} + 1}}{{{{10}^{2017}} + 1}}\] Giải thích các bước giải: Ta có: \(\begin{array}{l}A = \frac{{{{10}^{2019}} + 1}}{{{{10}^{2018}} + 1}} \Rightarrow \frac{A}{{10}} = \frac{{{{10}^{2019}} + 1}}{{{{10}^{2019}} + 10}} = \frac{{\left( {{{10}^{2019}} + 10} \right) – 9}}{{{{10}^{2019}} + 10}} = 1 – \frac{9}{{{{10}^{2019}} + 10}}\\B = \frac{{{{10}^{2018}} + 1}}{{{{10}^{2017}} + 1}} \Rightarrow \frac{B}{{10}} = \frac{{{{10}^{2018}} + 1}}{{{{10}^{2018}} + 10}} = \frac{{\left( {{{10}^{2018}} + 10} \right) – 9}}{{{{10}^{2018}} + 10}} = 1 – \frac{9}{{{{10}^{2018}} + 10}}\\{10^{2019}} > {10^{2018}} \Rightarrow {10^{2019}} + 10 > {10^{2018}} + 10 \Rightarrow \frac{9}{{{{10}^{2019}} + 10}} < \frac{9}{{{{10}^{2018}} + 10}} \Leftrightarrow \frac{{ – 9}}{{{{10}^{2019}} + 10}} > \frac{{ – 9}}{{{{10}^{2018}} + 10}}\\ \Rightarrow \frac{A}{{10}} > \frac{B}{{10}}\\ \Leftrightarrow A > B \Leftrightarrow \frac{{{{10}^{2019}} + 1}}{{{{10}^{2018}} + 1}} > \frac{{{{10}^{2018}} + 1}}{{{{10}^{2017}} + 1}}\end{array}\) Bình luận
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)
Đáp án:
\[\frac{{{{10}^{2019}} + 1}}{{{{10}^{2018}} + 1}} > \frac{{{{10}^{2018}} + 1}}{{{{10}^{2017}} + 1}}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = \frac{{{{10}^{2019}} + 1}}{{{{10}^{2018}} + 1}} \Rightarrow \frac{A}{{10}} = \frac{{{{10}^{2019}} + 1}}{{{{10}^{2019}} + 10}} = \frac{{\left( {{{10}^{2019}} + 10} \right) – 9}}{{{{10}^{2019}} + 10}} = 1 – \frac{9}{{{{10}^{2019}} + 10}}\\
B = \frac{{{{10}^{2018}} + 1}}{{{{10}^{2017}} + 1}} \Rightarrow \frac{B}{{10}} = \frac{{{{10}^{2018}} + 1}}{{{{10}^{2018}} + 10}} = \frac{{\left( {{{10}^{2018}} + 10} \right) – 9}}{{{{10}^{2018}} + 10}} = 1 – \frac{9}{{{{10}^{2018}} + 10}}\\
{10^{2019}} > {10^{2018}} \Rightarrow {10^{2019}} + 10 > {10^{2018}} + 10 \Rightarrow \frac{9}{{{{10}^{2019}} + 10}} < \frac{9}{{{{10}^{2018}} + 10}} \Leftrightarrow \frac{{ – 9}}{{{{10}^{2019}} + 10}} > \frac{{ – 9}}{{{{10}^{2018}} + 10}}\\
\Rightarrow \frac{A}{{10}} > \frac{B}{{10}}\\
\Leftrightarrow A > B \Leftrightarrow \frac{{{{10}^{2019}} + 1}}{{{{10}^{2018}} + 1}} > \frac{{{{10}^{2018}} + 1}}{{{{10}^{2017}} + 1}}
\end{array}\)