Question

[Viet Anh Hoang]

so sánh

10²⁰¹⁹-1/10²⁰²⁰-1 và 10²⁰¹⁸+1/10²⁰¹⁹+1

Answer 1

[Lê Tài Bảo Châu]

Áp dụng tính chất \(\frac{a}{b}< 1\Rightarrow\frac{a}{b}< \frac{a+m}{b+m}\left(m\in N\right)\)

Ta có: \(\frac{10^{2019}-1}{10^{2020}-1}< \frac{10^{2019}-1+11}{10^{2020}-1+11}=\frac{10^{2019}+10}{10^{2020}+10}=\frac{10.\left(10^{2018}+1\right)}{10.\left(10^{2019}+1\right)}=\frac{10^{2018}+1}{10^{2019}+1}\)

\(\Rightarrow\frac{10^{2019}-1}{10^{2020}-1}< \frac{10^{2018}+1}{10^{2019}+1}\)

Answer 2

[Huỳnh Quang Sang]

Đặt \(A=\frac{10^{2019}-1}{10^{2020}-1}\)

\(B=\frac{10^{2018}+1}{10^{2019}+1}\)

Dễ thấy \(A< 1\)

Áp dụng kết quả bài trên nếu \(\frac{a}{b}< 1\)thì \(\frac{a+m}{b+m}>\frac{a}{b}\)với m>0

Vậy \(A=\frac{10^{2019}-1}{10^{2020}-1}< \frac{\left[10^{2019}-1\right]+11}{\left[10^{2020}-1\right]+11}=\frac{10^{2019}+10}{10^{2020}+10}\)

\(A< \frac{10\left[10^{2018}+1\right]}{10\left[10^{2019}+1\right]}=\frac{10^{2018}+1}{10^{2019}+1}=B\)

Do đó : A<B