$(\sqrt6+\sqrt2)$.($\sqrt3-2$).$\sqrt{\sqrt3+2}$ 12/09/2021 Bởi Ariana $(\sqrt6+\sqrt2)$.($\sqrt3-2$).$\sqrt{\sqrt3+2}$
$(\sqrt{6}+\sqrt{2})(\sqrt{3}-2).\sqrt{\sqrt{3}+2}$ $=(3\sqrt{2}-2\sqrt{6}+\sqrt{6}-2\sqrt{2}).\sqrt{\sqrt{3}+2}$ $=(\sqrt{2}-\sqrt{6}).\sqrt{\sqrt{3}+2}$ $=\sqrt{2\sqrt{3}+4}-\sqrt{6\sqrt{3}+12}$ $=\sqrt{(1+2\sqrt{3}+3)}-\sqrt{(9+6\sqrt{3}+3)}$ $=\sqrt{(1+\sqrt{3}})^2-\sqrt{(3+\sqrt{3}})^2$ $=1+\sqrt{3}-3-\sqrt{3}$ $=-2$ Bình luận
$(\sqrt{6}+\sqrt{2})(\sqrt{3}-2).\sqrt{\sqrt{3}+2}$
$=(3\sqrt{2}-2\sqrt{6}+\sqrt{6}-2\sqrt{2}).\sqrt{\sqrt{3}+2}$
$=(\sqrt{2}-\sqrt{6}).\sqrt{\sqrt{3}+2}$
$=\sqrt{2\sqrt{3}+4}-\sqrt{6\sqrt{3}+12}$
$=\sqrt{(1+2\sqrt{3}+3)}-\sqrt{(9+6\sqrt{3}+3)}$
$=\sqrt{(1+\sqrt{3}})^2-\sqrt{(3+\sqrt{3}})^2$
$=1+\sqrt{3}-3-\sqrt{3}$
$=-2$
Đáp án:
Giải thích các bước giải: