Ta có x = 2 + \(\sqrt{3}\), do đó:
\(\left(x-1\right)\sqrt{3}=\left(2+\sqrt{3}-1\right)\sqrt{3}=\left(1+\sqrt{3}\right)\sqrt{3}\)
\(\sqrt{x^2-x+1}=\sqrt{\left(2+\sqrt{3}\right)^2-\left(2+\sqrt{3}\right)+1}=\sqrt{7+4\sqrt{3}-2-\sqrt{3}+1}=\sqrt{6+3\sqrt{3}}=\sqrt{3\left(2+\sqrt{3}\right)}=\sqrt{3}.\sqrt{2+\sqrt{3}}\)
Vậy M = \(\dfrac{\left(1+\sqrt{3}\right)\sqrt{3}}{\sqrt{3}.\sqrt{2+\sqrt{3}}}=\dfrac{1+\sqrt{3}}{\sqrt{2+\sqrt{3}}}=\dfrac{\left(1+\sqrt{3}\right)\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}}=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\left(1+\sqrt{3}\right)^2}}{\sqrt{\left(2+\sqrt{3}\right).\left(2-\sqrt{3}\right)}}=\dfrac{\sqrt{\left(2-\sqrt{3}\right).2\left(2+\sqrt{3}\right)}}{\sqrt{4-3}}=\sqrt{2\left(4-3\right)}=\sqrt{2}\)
