Question

So sánh A và B biết:

A=\(\dfrac{17^{18}+1}{17^{19}+1}\) , B=\(\dfrac{17^{17}+1}{17^{18}+1}\)

Answer 1

Tham khảo :

Answer 2

\(17A=\dfrac{17^{19}+17}{17^{19}+1}=\dfrac{\left(17^{19}+1\right)+16}{17^{19}+1}=\dfrac{17^{19}+1}{17^{19}+1}+\dfrac{16}{17^{19}+1}=1+\dfrac{16}{17^{19}+1}\)

\(17B=\dfrac{17^{18}+17}{17^{18}+1}=\dfrac{\left(17^{18}+1\right)+16}{17^{18}+1}=\dfrac{17^{18}+1}{17^{18}+1}+\dfrac{16}{17^{18}+1}=1+\dfrac{16}{17^{18}+1}\)

Vì \(17^{19}>17^{18}=>17^{19}+1>17^{18}+1\)

\(=>\dfrac{16}{17^{19}+1}< \dfrac{16}{17^{18}+1}\)

\(=>17A< 17B=>A< B\)