a) \(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}}\) = \(\dfrac{\sqrt{10+4\sqrt{6}}}{\sqrt{2}}+\dfrac{\sqrt{10-4\sqrt{6}}}{\sqrt{2}}\)
= \(\dfrac{\sqrt{\left(\sqrt{6}+2\right)^2}}{\sqrt{2}}+\dfrac{\sqrt{\left(\sqrt{6}-2\right)^2}}{\sqrt{2}}\) = \(\dfrac{\sqrt{6}+2}{\sqrt{2}}+\dfrac{\sqrt{6}-2}{\sqrt{2}}\)
= \(\dfrac{\sqrt{2}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{2}}+\dfrac{\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{2}}\) = \(\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}\)
= \(2\sqrt{3}\)
b) \(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\) = \(\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}-\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)
= \(\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}}-\dfrac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}}\) = \(\dfrac{\sqrt{3}-1}{\sqrt{2}}-\dfrac{\sqrt{3}+1}{\sqrt{2}}\)
= \(\dfrac{-2}{\sqrt{2}}\) = \(-\sqrt{2}\)
\(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}}=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}+1+\sqrt{5}-1=2\sqrt{5}\)
b) Đặt \(B=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
\(B\sqrt{2}=\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}\)
\(B\sqrt{2}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}-1-\sqrt{3}-1=-2\)
\(B\sqrt{2}=-2\Rightarrow B=\dfrac{-2}{\sqrt{2}}=-\sqrt{2}\)
Sửa câu a:
\(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}}=\sqrt{\left(\sqrt{3}+\sqrt{2}^2\right)}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\)
