cho A= 196/197+197/198; B=196+197/197+198 Trong hai sao A và B, số nào lớn hơn?
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\(\dfrac{196}{197+198}< \dfrac{196}{197};\dfrac{197}{197+198}< \dfrac{197}{198}\)
=>B<A
\(B=\frac{196+197}{197+198}=\frac{196}{197+198}+\frac{197}{197+198}\)
\(\frac{196}{197}>\frac{196}{197+198};\frac{197}{198}>\frac{197}{197+198}\)
=>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}
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=196/197+197/198>196/197+198 +197/197+198= 196+197/197+198=B
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
`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`