Áp dụng tính chất \(\dfrac{a}{b}< 1\Rightarrow\dfrac{a}{b}< \dfrac{a+m}{b+m}\) ta có:
\(A=\dfrac{10^{1992}+1}{10^{1993}+1}< \dfrac{10^{1992}+1+9}{10^{1993}+1+9}=\dfrac{10^{1992}+10}{10^{1993}+10}\)
\(=\dfrac{10\left(10^{1991}+1\right)}{10\left(10^{1992}+1\right)}=\dfrac{10^{1991}+1}{10^{1992}+1}\)
\(\Rightarrow\dfrac{10^{1992}+1}{10^{1993}+1}< \dfrac{10^{1991}+1}{10^{1992}+1}\)
Hay \(A>B\)
Ta đi so sánh:
\(\dfrac{1}{A}=\dfrac{10^{1991}+1}{10^{1992}+1}\) và \(\dfrac{1}{B}=\dfrac{10^{1992}+1}{10^{1993}+1}\)
Ta có:\(\dfrac{10}{A}=\dfrac{10^{1992}+10}{10^{1992}+1}\)
\(=\dfrac{10^{1992}+1+9}{10^{1992}+1}\)
\(=1+\dfrac{9}{10^{1992}+1}\)
\(\dfrac{10}{B}=\dfrac{10^{1993}+10}{10^{1993}+1}\)
\(=\dfrac{10^{1993}+1+9}{10^{1993}+11}\)
Mà \(\dfrac{9}{10^{1992}+1}>\dfrac{9}{10^{1993}+1}\)
\(\Rightarrow\dfrac{10}{A}>\dfrac{10}{B}\)
Vậy A<B
