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\(B=\frac{196+197}{197+198}=\frac{196}{197+198}+\frac{197}{197+198}\)
\(\frac{196}{197}>\frac{196}{197+198};\frac{197}{198}>\frac{197}{197+198}\)
=>A>B
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
\(A=\frac{196}{197}+\frac{197}{198}=\left(1-\frac{1}{197}\right)+\left(1-\frac{1}{198}\right)=1-\frac{1}{197}+1-\frac{1}{198}=1-\frac{1}{197}+\frac{197}{197}-\frac{1}{198}\)\(=1-\frac{198}{197}-\frac{1}{198}=\frac{197}{197}-\frac{198}{197}-\frac{1}{198}=\frac{-1}{197}-\frac{1}{198}<\frac{196+197}{197+198}=\frac{393}{395}\)
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 \(\frac{197}{198}>\frac{197}{197+198}\)
\(\Rightarrow A>B\)
A = 192.198 = 192.(200 - 2) = 192.200 - 192.2 = 192.200 - 384
B = 193.197 = 193.(200 - 3) = 193.200 - 193.3 = (192 + 1).200 - 193.3 = 192.200 + 1.200 - 193.3 = 192.200 - 379
Vì 384 > 379 nên 192.200 - 384 < 192.200 - 379 hay A < B
Vậy A < B.