Cho:
A=\(\dfrac{196}{197}+\dfrac{197}{198}\) và B=\(\dfrac{196+197}{197+198}\) .Trong 2 số, số nào lớn hơn?
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Ta có:
\(A=\dfrac{196}{197}+\dfrac{197}{198}\)
\(B=\dfrac{196+197}{197+198}\)
\(=\dfrac{196}{197+198}+\dfrac{197}{197+198}\)
Áp dụng tính chất \(\dfrac{a}{b}>\dfrac{a}{b+m}\) ta có:
\(\left\{{}\begin{matrix}\dfrac{196}{197}>\dfrac{196}{197+198}\\\dfrac{197}{198}>\dfrac{197}{197+198}\end{matrix}\right.\)
\(\Rightarrow\dfrac{196}{197}+\dfrac{197}{198}>\dfrac{196}{197+198}+\dfrac{197}{197+198}=\dfrac{196+197}{197+198}\)
Vậy \(A>B\)
vì \(\dfrac{196}{197+198}< \dfrac{196}{197};\dfrac{197}{197+198}< \dfrac{197}{198}\)
nên \(A=\dfrac{196}{197}+\dfrac{197}{198}>\dfrac{196}{197+198}+\dfrac{197}{197+198}=\dfrac{196+197}{197+198}=B\)
=>A>B
vậy....
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
=>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 : \(\frac{196}{197}>\frac{196}{198}+\frac{197}{198}=\frac{196+197}{198}>\frac{196+197}{197+198}\)
\(\Rightarrow A>B\)
\(B=\frac{196+197}{197+198}=\frac{196}{197+198}+\frac{197}{197+198}\)
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
\(\dfrac{196}{197}>\dfrac{196}{197+198};\dfrac{197}{198}>\dfrac{197}{197+198}\)
nên A>B
Ta phân tích số B
B = 196 + 197/197 + 198 = 196 + 197/395 = 196/395 + 197/395
Ta thấy
196/197 > 196/395
197/198 > 197/395
=> A > B
Vậy A > B