Question

Cho: A =(196/197)+(197/198)

B=(196+197/197+198)

So sánh A&B

Answer 1

Dễ thấy \(B=\dfrac{196+197}{197+198}< 1\)

Ta có:

\(A=\dfrac{196}{197}+\dfrac{197}{198}\)

\(A=1-\dfrac{1}{197}+1-\dfrac{1}{198}\)

\(A=\left(1+1\right)-\left(\dfrac{1}{197}+\dfrac{1}{198}\right)\)

\(A=2+\left(\dfrac{1}{197}+\dfrac{1}{198}\right)\)

Mà \(\left(\dfrac{1}{197}+\dfrac{1}{198}\right)< 1\) vì \(\dfrac{1}{197}< 0,5;\dfrac{1}{198}< 0,5\) nên \(\left(\dfrac{1}{197}+\dfrac{1}{198}\right)< 1\)

\(\Rightarrow2-\left(\dfrac{1}{197}+\dfrac{1}{198}\right)>1\)

\(\Rightarrow A>1\)

Vì \(A>1;B< 1\) nên \(A>B\).