Question

Rút gọn biểu thức:

a, A= \(\sqrt{2-\sqrt{3}}\left(\sqrt{6}-\sqrt{2}\right)\left(2+\sqrt{3}\right)\)

b, \(\frac{\left(2+\sqrt{3}\right)\sqrt{2-\sqrt{3}}}{\sqrt{2}+\sqrt{3}}\)

Giúp mih nhanh với ạ. Cần gấp! Cảm ơn nhiều ạ

Answer 1

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

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

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

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

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

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

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

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

A=3-1

A=2

Vậy A=2

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