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

`Answer:`

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

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

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

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

\(=2-\left(\frac{1}{197}+\frac{1}{198}\right)\)

Mà \(\frac{1}{197}< \frac{1}{2}\) và \(\frac{1}{198}< \frac{1}{2}\)

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

\(\Rightarrow2-\left(\frac{1}{197}+\frac{1}{198}\right)>1\) hay \(A>1\)

Mà \(B=\frac{196+197}{197+198}=\frac{393}{395}< 1\)

`=>A>B`