Question

\(\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-2\right)\sqrt{\sqrt{3}+2}\)

Answer 1

rút gọn nha các bạn. Giúp mk vs

Answer 2

\(\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-2\right)\sqrt{\sqrt{3}+2}\\ =\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right)\sqrt{2\sqrt{3}+4}\\ =\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right)\sqrt{2\sqrt{3}+3+1}\\ =\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right)\sqrt{\left(\sqrt{3}+1\right)^2}\\ =\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right)\left(\sqrt{3}+1\right)\\ =\left(\sqrt{3}-2\right)\left(\sqrt{3}+1\right)^2\\ =\left(\sqrt{3}-2\right)\left(3+2\sqrt{3}+1\right)\\ =\left(\sqrt{3}-2\right)\left(4+2\sqrt{3}\right)\\ =2\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)\\ =2\left(3-4\right)=-2\)