Question

Cho: A = \(\frac{10^{2001}+1}{10^{2002}+1}\) và B = \(\frac{10^{2002}+1}{10^{2003}+1}\)

Hãy so sánh A và B.

Answer 1

\(A=\frac{10^{2001}+1}{10^{2002}+1}=\frac{\left(10^{2001}+1\right)\left(10^{2003}+1\right)}{\left(10^{2002}+1\right)\left(10^{2003}+1\right)}=\frac{10^{4004}+10^{2001}+10^{2003}+1}{\left(10^{2002}+1\right)\left(10^{2003}+1\right)}\)

\(B=\frac{10^{2002}+1}{10^{2003}+1}=\frac{\left(10^{2002}+1\right)\left(10^{2002}+1\right)}{\left(10^{2003}+1\right)\left(10^{2002}+1\right)}=\frac{10^{4004}+2.10^{2002}+1}{\left(10^{2003}+1\right)\left(10^{2002}+1\right)}\)

Vì 10²⁰⁰¹ + 10²⁰⁰³ < 2.10²⁰⁰² nên A < B

Answer 2

Không nhầm là quy đồng phân số A nhân với 10

Answer 3

Đặt A = \(\frac{10^{2001}+1}{10^{2002}+1}\) ; B = \(\frac{10^{2002}+1}{10^{2003}+1}\)

10A = \(\frac{10\left(10^{2001}+1\right)}{10^{2002}+1}\) \(=\frac{10^{2002}+1+9}{10^{2002}+1}\)

10B = \(y=\frac{10\left(10^{2002}+1\right)}{10^{2013}+1}\)\(=\frac{10^{2003}+1+9}{10^{2003}+1}\)

Vì \(y=\frac{10^{2002}+1+9}{10^{2002}+1}\) > \(\frac{10^{2003}+1+9}{10^{2003}+1}\) nên 10A > 10B nên A > B

Cũng ko chắc nữa -_-