Đáp án:
$C=1$
Giải thích các bước giải:
$C=\left(4-\sqrt{15}\right)\left(\sqrt{2-\sqrt 3}+\sqrt{3+\sqrt 5}\right)^2$
$=\dfrac{4-\sqrt{15}}{2}.2.\left(\sqrt{2-\sqrt 3}+\sqrt{3+\sqrt 5}\right)^2$
$=\dfrac{4-\sqrt{15}}{2}.\left(\sqrt 2.\sqrt{2-\sqrt 3}+\sqrt 2.\sqrt{3+\sqrt 5}\right)^2$
$=\dfrac{4-\sqrt{15}}{2}.\left(\sqrt{4-2\sqrt 3}+\sqrt{6+2\sqrt 5}\right)^2$
$=\dfrac{4-\sqrt{15}}{2}.\left(\sqrt{3-2\sqrt 3+1}+\sqrt{5+2\sqrt 5+1}\right)^2$
$=\dfrac{4-\sqrt{15}}{2}.\left[\sqrt{(\sqrt 3-1)^2}+\sqrt{(\sqrt 5+1)^2}\right)^2$
$=\dfrac{4-\sqrt{15}}{2}.\left(|\sqrt 3-1|+|\sqrt 5+1|\right)^2$
$=\dfrac{4-\sqrt{15}}{2}.\left(\sqrt 3-1+\sqrt 5+1\right)^2$
$=\dfrac{4-\sqrt{15}}{2}.\left(\sqrt 3+\sqrt 5\right)^2$
$=\dfrac{4-\sqrt{15}}{2}.\left(8+2\sqrt{15}\right)$
$=\left(4-\sqrt{15}\right).\left(4+\sqrt{15}\right)$
$=16-15$
$=1$.
