\(A=\sqrt{625}-\dfrac{1}{\sqrt{5}}=25-\dfrac{1}{\sqrt{5}}\)
\(B=\sqrt{576}-\dfrac{1}{\sqrt{6}}+1=24-\dfrac{1}{\sqrt{6}}+1=25-\dfrac{1}{\sqrt{6}}.\)
Vì \(\sqrt{5}< \sqrt{6}\) nên \(\dfrac{1}{\sqrt{5}}>\dfrac{1}{\sqrt{6}}.\)
Từ (1), (2) và (3) suy ra \(A< B.\)
A=\(\sqrt{625}\)−\(\dfrac{1}{\sqrt{5}}\)
⇒A= 25-\(\dfrac{1}{\sqrt{5}}\)
B =\(\sqrt{576}\) - \(\dfrac{1}{\sqrt{6}}+1\)
⇒B = 24-\(\dfrac{1}{\sqrt{6}}+1\)
Hay: B = (24+1)-\(\dfrac{1}{\sqrt{6}}\)
⇒ B=25-\(\dfrac{1}{\sqrt{6}}\)
Vì: 25-\(\dfrac{1}{\sqrt{5}}\) > 25-\(\dfrac{1}{\sqrt{6}}\)
Vậy: A > B