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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 \(A>B\)
\(\dfrac{196}{197}>\dfrac{196}{197+198};\dfrac{197}{198}>\dfrac{197}{197+198}\)
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
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
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....