Question

Giải giúp em ạ
So sánh A = 2√3 và B= \(\sqrt{4+2\sqrt{3}}\)
A= \(\sqrt{5+2\sqrt{6}}\) và B = \(\dfrac{2\sqrt{2}}{1+2\text{√2}-\sqrt{3}}\)

Answer 1

a: \(B=\sqrt{3}+1< \sqrt{3}+\sqrt{3}=A\)

b: \(A=\sqrt{3}+\sqrt{2}\)

\(B=\dfrac{2\sqrt{2}\left(1+2\sqrt{2}+\sqrt{3}\right)}{9+4\sqrt{2}-3}=\dfrac{2\sqrt{2}\left(1+2\sqrt{2}+\sqrt{3}\right)}{6+4\sqrt{2}}\)

\(=\dfrac{\sqrt{2}\left(2\sqrt{2}+\sqrt{3}+1\right)}{3+2\sqrt{2}}\)

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

\(=\left(3\sqrt{2}-4\right)\left(2\sqrt{2}+\sqrt{3}+1\right)\)

\(=12+3\sqrt{6}+3\sqrt{2}-8\sqrt{2}-4\sqrt{3}-4\)

\(=8+3\sqrt{6}-5\sqrt{2}-4\sqrt{3}< \sqrt{3}+\sqrt{2}\)