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 \(A>B\)

\(\dfrac{196}{197}>\dfrac{196}{197+198};\dfrac{197}{198}>\dfrac{197}{197+198}\)

nên A>B

Ta phân tích số B

B = 196 + 197/197 + 198 = 196 + 197/395 = 196/395 + 197/395

Ta thấy

196/197 > 196/395

197/198 > 197/395

=> A > B

Vậy A > B

bằng nhau mà

bằng nhau mà ???

 Ta có : \(A=\frac{196}{197}+\frac{197}{198}\)   ;  \(B=\frac{196+197}{197+198}\)\(=\frac{196}{197+198}\) \(+\) \(\frac{197}{197+198}\)

Ta thấy :

\(\frac{196}{197}>\frac{196}{197+198}\)

\(\frac{197}{198}>\frac{197}{197+198}\)

\(\frac{\Rightarrow196}{197}+\frac{197}{198}>\frac{196}{197+198}+\frac{197}{197+198}\)

hay A>B

\(B=\frac{196+197}{197+198}=\frac{196}{197+198}+\frac{197}{197+198}\)

Ta có \(\frac{196}{197}>\frac{196}{197+198}\)và\(\frac{197}{198}>\frac{197}{197+198}\)

=>\(\frac{196}{197}+\frac{197}{198}>\frac{196+197}{197+198}\)

=>A>B

A<B

k nha     Lê Nguyên Khánh

\(A=\frac{196}{197}+\frac{197}{198}>\frac{196}{197+198}+\frac{197}{197+198}=\frac{196+197}{197+198}=B\)

Vậy \(A>B\)

B > A vì 196/197<197/197,197/198<196+198

B=393/395=1-(2/395)=1-[(1/395)+(1/395)]   {2}

A=1-(1/197)-(1/198)=1-[(1/197)+(1/198)]   {1}

vÌ {1} >{2}

=>A>B

A=196/197+197/198>196/197+198 +197/197+198= 196+197/197+198=B