\(A=\sqrt{11+\sqrt{96}}=\sqrt{\left(2\sqrt{2}+\sqrt{3}\right)^2}=2\sqrt{2}+\sqrt{3}\)

\(B=\frac{2\sqrt{2}}{1+\sqrt{2}-\sqrt{3}}=\frac{2\sqrt{2}\left(1+\sqrt{2}+\sqrt{3}\right)}{2\sqrt{2}}=1+\sqrt{2}+\sqrt{3}\)

Xét : \(A-B=2\sqrt{2}+\sqrt{3}-\left(1+\sqrt{2}+\sqrt{3}\right)=\sqrt{2}-1>0\)

\(\Rightarrow A>B\)

\(B=\frac{2\sqrt{2}}{1+\sqrt{2}-\sqrt{3}}=\frac{2\sqrt{2}\left(1+\sqrt{2}+\sqrt{3}\right)}{\left(1+\sqrt{2}-\sqrt{3}\right)\left(1+\sqrt{2}+\sqrt{3}\right)}.\)\(=\frac{2\sqrt{2}\left(1+\sqrt{2}+\sqrt{3}\right)}{\left(1+\sqrt{2}\right)^2-3}=1+\sqrt{2}+\sqrt{3}\)

\(A=\sqrt{11+\sqrt{96}}=\sqrt{11+4\sqrt{6}}=\sqrt{8+2.2\sqrt{2}.\sqrt{3}+3}=\sqrt{\left(2\sqrt{2}+\sqrt{3}\right)^2}\)\(=2\sqrt{2}+\sqrt{3}>1+\sqrt{2}+\sqrt{3}=B\)

a)So sánh vs 5/2

b)So sánh vs 40/9

a) \(2-2\sqrt{3}\) và \(4-\sqrt{15}\)

Giả sử : \(2-2\sqrt{3}\ge4-\sqrt{15}\)

⇔ \(\sqrt{15}-2\sqrt{3}\ge2\)

⇔ \(\left(\sqrt{15}-2\sqrt{3}\right)^2\ge2^2\)

⇔ 15 - \(12\sqrt{5}+12\) ≥ 4

⇔ 27 -4 ≥ \(12\sqrt{5}\)

⇔ 23 ≥ \(12\sqrt{5}\)

⇔ \(23^2\) ≥ \(\left(12\sqrt{5}\right)^2\)

⇔ 529 ≥ 720 (sai)

Vậy 2 - \(2\sqrt{3}< 4-\sqrt{15}\)

b) \(\sqrt{11}+2\) và \(3+\sqrt{3}\)

Giả sử : \(\sqrt{11}+2\le3+\sqrt{3}\)

⇔ \(\sqrt{11}-\sqrt{3}\le1\)

⇔ \(\left(\sqrt{11}-\sqrt{3}\right)^2\le1\)

⇔ 14 - \(2\sqrt{33}\) ≤ 1

⇔ 13 ≤ \(2\sqrt{33}\)

⇔ \(13^2\le\left(2\sqrt{33}\right)^2\)

⇔ 169 ≤ 132 (sai)

Vậy \(\sqrt{11}+2\ge3+\sqrt{3}\)

Nguyễn Thanh Hằng, Dương Nguyễn, Ngô Thành Chung, Khôi Bùi , Trần Nguyễn Bảo Quyên, Tạ Thị Diễm Quỳnh, Nguyễn Quang Minh, Khánh Như Trương Ngọc, Nguyễn Quang Minh, Mysterious Person, Phùng Khánh Linh, JakiNatsumi, DƯƠNG PHAN KHÁNH DƯƠNG, Hoàng Phong, Ribi Nkok Ngok, ...