Đề mk chắc là sai đó bn !!

Bn kiểm tra lại xem

Nhớ đúng mk nha

=\(\frac{198.199-200}{198.197-200}\)

=\(\frac{1.199-200}{1.197-200}\)

=\(\frac{199-200}{197-200}\)

=\(\frac{-1}{-3}\)

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

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

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

=> A>B

a>b nhé t

Ta có :

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

Vì : 

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

=> A > B
=))

ta có:

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

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

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

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

Vậy:\(A>B\)

A>1                                       B<1

     Vậy A>B

=1 nha bạn

198*199-200/196+197*198

=1

OK

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 \(A>B\)

= đó bạn

= 1  đó bạn

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

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

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

Vậy : \(A>B\)

\(A>B\)

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

=>A>B

vậy...