Chứng minh rằng: (5 +2 căn bậc hai của 6 / căn bậc hai của 3

* Hướng dẫn giải

Nhận xét rằng:

$\begin{array}{l}5+2\sqrt{6}={\left(\sqrt{3}\right)}^2+2\sqrt{3}\sqrt{2}+{\left(\sqrt{2}\right)}^2={\left(\sqrt{3}+\sqrt{2}\right)}^2\\ 5-2\sqrt{6}={\left(\sqrt{3}\right)}^2-2\sqrt{3}\sqrt{2}+{\left(\sqrt{2}\right)}^2={\left(\sqrt{3}-\sqrt{2}\right)}^2\end{array}$

Suy ra:

${\left(\frac{5+2\sqrt{6}}{\sqrt{3}+\sqrt{2}}\right)}^2={\left(\frac{{\left(\sqrt{3}+\sqrt{2}\right)}^2}{\sqrt{3}+\sqrt{2}}\right)}^2={\left(\sqrt{3}+\sqrt{2}\right)}^2=5+2\sqrt{6}$

${\left(\frac{5-2\sqrt{6}}{\sqrt{3}-\sqrt{2}}\right)}^2={\left(\frac{{\left(\sqrt{3}-\sqrt{2}\right)}^2}{\sqrt{3}-\sqrt{2}}\right)}^2={\left(\sqrt{3}-\sqrt{2}\right)}^2=5-2\sqrt{6}$

Do đó:

$VT=5+2\sqrt{6}-(5-2\sqrt{6})=4\sqrt{6}$