Question

Tính

A=\(\dfrac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}\)

B=\(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

C=\(\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\)

Answer 1

a:\(A=\dfrac{6\sqrt{2}-4\sqrt{3}+2\sqrt{5}}{6\sqrt{2}-6\sqrt{3}+3\sqrt{5}}=\dfrac{2\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}{3\left(2\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}\)

b: \(B=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)

c: \(C=\sqrt{6-2\sqrt{5}}\cdot\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\)

\(=\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)\)

\(=18+6\sqrt{5}-6\sqrt{5}-10=8\)