Bài 2:
\(A=\sqrt{26}+\sqrt{10}>\sqrt{25}+\sqrt{9}=5+3=8\)
\(B=\sqrt{64}=8\)
Do đó: A>B
1.Ta có:
\(A=\)\(\sqrt{13}+\sqrt{20}=\sqrt{13}+2\sqrt{5}\)
\(B=\)\(\sqrt{24}+\sqrt{19}=\sqrt{19}+2\sqrt{6}\)
So sánh ta thấy:
\(\sqrt{13}<\sqrt{19}\) ; \(2\sqrt{5}<2\sqrt{6}\)
Vậy A < B