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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
\(A=\frac{196}{197}+\frac{197}{198}>\frac{196}{197+198}+\frac{197}{197+198}=\frac{196+197}{197+198}=B\)
Vậy \(A>B\)
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`
=>Ta có: B=\(\frac{196}{197+198}\) + \(\frac{197}{197+198}\)\
Ta có: \(\frac{196}{197+198}\) < \(\frac{196}{197}\) ; \(\frac{197}{197+198}\)< \(\frac{197}{198}\)
=> A= \(\frac{196}{197}\)+ \(\frac{197}{198}\) > \(\frac{196}{197+198}\) + \(\frac{197}{197+198}\) = B. Vậy A>B
Ta có : \(\frac{196}{197}>\frac{196}{198}+\frac{197}{198}=\frac{196+197}{198}>\frac{196+197}{197+198}\)
\(\Rightarrow A>B\)
bằng nhau mà
bằng nhau mà ???