Question

[Nguyễn Thảo An]

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

và B=\(^{\frac{10^{2002}+_{ }1}{10^{2003}+1}}\)

So sánh A và B

Answer 1

[Nguyễn Quỳnh Chi]

Ta c/m bài toán phụ:

Giả sử a<b (a,b\(\in\)N; b\(\ne\)0)

So sánh \(\frac{a}{b}\) với \(\frac{a+m}{b+m}\) (m\(\in\)N*)

Có: \(\frac{a}{b}=\frac{a\left(b+m\right)}{b\left(b+m\right)}=\frac{ab+am}{b\left(b+m\right)}\)

\(\frac{a+m}{b+m}=\frac{b\left(a+m\right)}{b\left(b+m\right)}=\frac{ab+bm}{b\left(b+m\right)}\)

Vì a<b \(\Rightarrow\) am<bm (m\(\in\)N*) \(\Rightarrow\) ab+am<ab+bm

\(\Rightarrow\frac{ab+am}{b\left(b+m\right)}< \frac{ab+bm}{b\left(b+m\right)}\) hay \(\frac{a}{b}< \frac{a+m}{b+m}\)

Áp dụng bài toán trên ta có:

\(B=\frac{10^{2002}+1}{10^{2003}+1}< \frac{10^{2002}+1+9}{10^{2003}+1+9}=\frac{10^{2002}+10}{10^{2003}+10}=\frac{10\left(10^{2001}+1\right)}{10\left(10^{2002}+1\right)}=\frac{10^{2001}+1}{10^{2002}+1}=A\)

\(\Rightarrow B< A\)

Vậy B<A