A1. What do we get after rationalizing the expression $\frac{4+2 \sqrt{6}}{(1+\sqrt{2})+\sqrt{3}}$?
(A) $1-\sqrt{2}-\sqrt{3}$
(B) $1-\sqrt{2}+\sqrt{3}$
(C) $1+\sqrt{2}+\sqrt{3}$
(D) $-1+\sqrt{2}+\sqrt{3}$
(E) $-1+\sqrt{2}-\sqrt{3}$
A1. The numerator and denominator are multiplied by the expression $(1+\sqrt{2})-\sqrt{3}$ and we get $\frac{4+2 \sqrt{6}}{(1+\sqrt{2})+\sqrt{3})}$. $\frac{(1+\sqrt{2})-\sqrt{3}}{(1+\sqrt{2})-\sqrt{3}}=\frac{4+4 \sqrt{2}-4 \sqrt{3}+2 \sqrt{6}+2 \sqrt{12}-2 \sqrt{18}}{(1+\sqrt{2})^{2}-3}=\frac{4-2 \sqrt{2}+2 \sqrt{6}}{2 \sqrt{2}}=\frac{2-\sqrt{2}+\sqrt{6}}{\sqrt{2}}$. The obtained expression is rationalized with $\sqrt{2}$ and simplified by $2$, and we get the result $-1+\sqrt{2}+\sqrt{3}$.