Question

Tính:

a) \(\dfrac{\sqrt{3-\sqrt{5}}.\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)

b) \(\dfrac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)

c) \(\dfrac{\left(\sqrt{5}+2\right)^2-8\sqrt{5}}{2\sqrt{5}-4}\)

Mọi người giúp em với! Em cám ơn trước ạ.

Answer 1

a. Ta có:A = \(\dfrac{\sqrt{3-\sqrt{5}}.\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)

= \(\dfrac{\sqrt{3-\sqrt{5}}.\left(\sqrt{3+\sqrt{5}}\right)^2}{\sqrt{2}.\left(\sqrt{5}-1\right)}\)

= \(\dfrac{\sqrt{9-5}.\left(\sqrt{3}+\sqrt{5}\right)}{\sqrt{2}.\left(\sqrt{5}-1\right)}\)

= \(\dfrac{2\sqrt{3+\sqrt{5}}}{\sqrt{2}.\left(\sqrt{5+1}\right)}\)

=\(\dfrac{\sqrt{2}.\sqrt{3+\sqrt{5}}}{\sqrt{5}+1}\)

⇒A² = \(\dfrac{2.\left(3+\sqrt{5}\right)}{5+2\sqrt{5}+1}\)=\(\dfrac{6+2\sqrt{5}}{6+2\sqrt{5}}\)=1

⇒A=\(\sqrt{1}\)=1