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

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

=>A>B

\(A=\frac{196}{197}+\frac{197}{198}=\left(1-\frac{1}{197}\right)+\left(1-\frac{1}{198}\right)=1-\frac{1}{197}+1-\frac{1}{198}=1-\frac{1}{197}+\frac{197}{197}-\frac{1}{198}\)\(=1-\frac{198}{197}-\frac{1}{198}=\frac{197}{197}-\frac{198}{197}-\frac{1}{198}=\frac{-1}{197}-\frac{1}{198}

Tui cx k bt

\(=1-\frac{198}{197}-\frac{1}{198}=\frac{197}{197}-\frac{198}{197}-\frac{1}{198}=\frac{-1}{197}-\frac{1}{198}

A>B chắc luôn likeeeeeeeeeeeeeeeeeee đi

takefusa kubo

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

ta có \(\frac{196}{197+198}\)< \(\frac{196}{197}\)

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

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

=> B < A

A=b ngu nhu cho

A bằng b nha bạn 

A=B , K CHO MINK NHÉ

Vi 196/197 =196/197

=>ta so sanh 197/198 va 198

198=39204/198

197/198 < 39204/198 (vi 197 < 39204 )

=>A < B 

Vay A < B 

MÌH LÀ NG THANKS BẠN MỚI ĐÚG THANKS

không hiểu lắm

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

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

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

Vậy A > B

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

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

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

\(\Rightarrow A>B\)

Ta có:

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

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

\(\Rightarrow\frac{196}{197+198}+\frac{197}{197+198}<\frac{196}{197}+\frac{197}{198}\left(1\right)\)

Mà \(B=\frac{196+197}{197+198}=\frac{196}{197+198}+\frac{197}{197+198}\left(2\right)\)

Từ ( 1 ) và ( 2 )

\(\Rightarrow B<\frac{196}{197}+\frac{197}{198}\)

\(\Rightarrow B

Gọi 196/197 là a/c, 197/198 la b/d

Có a/c+b/d=a+b/c+d( tính chất của dãy tỉ số bằng nhau)

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

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

Ta có: \(\frac{196}{197}>\frac{196}{197+198}\) vì \(197<197+198\)

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

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

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

Vậy \(A>B\)