Đặt 4453=a; 1997=b
\(A=\left(5+\dfrac{6}{a}\right)\cdot\dfrac{1}{b}-\dfrac{2}{b}\cdot\left(2+\dfrac{3}{a}\right)\)
\(=\dfrac{5a+6}{a}\cdot\dfrac{1}{b}-\dfrac{2}{b}\cdot\dfrac{2a+3}{a}\)
\(=\dfrac{5a+6-4a-6}{ab}=\dfrac{a}{ab}=\dfrac{1}{b}=\dfrac{1}{1997}\)