\(A=\dfrac{25^{10}+1}{25^{10}-1}=\dfrac{25^{10}-1+2}{25^{10}-1}=\dfrac{25^{10}-1}{25^{10}-1}+\dfrac{2}{25^{10}-1}=1+\dfrac{2}{25^{10}-1}\left(1\right)\)
\(B=\dfrac{25^{10}-1}{25^{10}-3}=\dfrac{25^{10}-3+2}{25^{10}-3}=\dfrac{25^{10}-3}{25^{10}-3}+\dfrac{2}{25^{10}-3}=1+\dfrac{2}{25^{10}-3}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow A< B\)
Ta có : \(\dfrac{25^{10}+1}{25^{10}-1}=\dfrac{25^{10}-1+2}{25^{10}-1}=\dfrac{25^{10}-1}{25^{10}-1}+\dfrac{2}{25^{10}-1}\)\(=1+\dfrac{2}{25^{10}-1}\)
Ta có : \(\dfrac{25^{10}-1}{25^{10}-3}=\dfrac{25^{10}-3+2}{25^{10}-3}=\dfrac{25^{10}-3}{25^{10}-3}+\dfrac{2}{25^{10}-3}=1+\dfrac{2}{25^{10}-3}\)
Vì \(25^{10}-1>25^{10}-3\)
\(\Rightarrow\dfrac{2}{25^{10}-1}< \dfrac{2}{25^{10}-3}\)
\(\Rightarrow1+\dfrac{2}{25^{10}-1}< 1+\dfrac{2}{25^{10}-3}\)
\(\Rightarrow A< B\)
Vậy \(A< B\)
