Question

So sánh : A =\(\dfrac{17^{18}+1}{17^{19}+1}\) và B =\(\dfrac{17^{17}+1}{17^{18}+1}\)

Answer 1

Ta có: \(17A=\frac{17^{19}+17}{17^{19}+1}=1+\frac{16}{17^{19}+1}\)

\(17B=\frac{17^{18}+17}{17^{18}+1}=1+\frac{16}{17^{18}+1}\)

Vì \(\frac{16}{17^{19}+1}>\frac{16}{17^{18}+1}\Rightarrow1+\frac{16}{17^{19}+1}>1+\frac{16}{17^{18}+1}\)

\(\Rightarrow17A>17B\)

\(\Rightarrow A>B\)

Vậy A > B

Answer 2

A < B

Answer 3

\(A=\dfrac{17^{18}+1}{17^{19}+1}=\dfrac{17^{18}+1}{17\left(17^{18}+1\right)}=\dfrac{1}{17}\)

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

Vậy \(A=B\left(=\dfrac{1}{17}\right)\).

Answer 4