Question

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

b)\(\frac{\left(\sqrt{a}-1\right)\left(\sqrt{6}-\sqrt{2}\right)\left(a-\sqrt{ab}\right)}{\left(a\sqrt{a}-a\right)\left(a-b\right)}\) (Với a,b >0 và a khác 1)

Answer 1

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

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

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

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

\(B=\frac{\left(\sqrt{a}-1\right)\left(\sqrt{6}-\sqrt{2}\right)\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}{a\left(\sqrt{a}-1\right)\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\frac{\sqrt{6}-\sqrt{2}}{a+\sqrt{ab}}\)