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
