Question

$$(\frac{3+12}{\sqrt{3},2})^{2}+\boxed{\frac{1}{(\sqrt{3}+\sqrt{2})}}$$

Answer

$$(3+12*sqrt(3),2)^2+boxe*(sqrt(3)+sqrt(2))*d$$

Solution


Simplify  \(boxed\times 1\times (\sqrt{3}+\sqrt{2})\)  to  \(dboxe(\sqrt{3}+\sqrt{2})\).
\[{(3+12\sqrt{3},2)}^{2}+dboxe(\sqrt{3}+\sqrt{2})\]
Regroup terms.
\[{(3+12\sqrt{3},2)}^{2}+boxe(\sqrt{3}+\sqrt{2})d\]