Question

Rút gọn biểu thức

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

Answer 1

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

Answer 2

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

\(=\sqrt{2}.\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right).\sqrt{2+\sqrt{3}}\)

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

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

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

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

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

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

\(=-2\)