Ta có:
\(A=\dfrac{196}{197}+\dfrac{197}{198}\)
\(B=\dfrac{196+197}{197+198}\)
\(=\dfrac{196}{197+198}+\dfrac{197}{197+198}\)
Áp dụng tính chất \(\dfrac{a}{b}>\dfrac{a}{b+m}\) ta có:
\(\left\{{}\begin{matrix}\dfrac{196}{197}>\dfrac{196}{197+198}\\\dfrac{197}{198}>\dfrac{197}{197+198}\end{matrix}\right.\)
\(\Rightarrow\dfrac{196}{197}+\dfrac{197}{198}>\dfrac{196}{197+198}+\dfrac{197}{197+198}=\dfrac{196+197}{197+198}\)
Vậy \(A>B\)