a) Ta có:
$10 + \sqrt{60} -\sqrt{24} - \sqrt{40}$
$= 5 + 3 + 2 + 2\sqrt{15} - 2\sqrt{6} - 2\sqrt{10}$
$= (\sqrt5)^2 + (\sqrt3)^2 + (\sqrt2)^2 + 2\sqrt3.\sqrt5 - 2\sqrt2.\sqrt3 - 2\sqrt2.\sqrt5$
$= (\sqrt2 -\sqrt5 - \sqrt3)^2$
$= (\sqrt3 + \sqrt5 - \sqrt2)^2$
Do đó:
$\sqrt{10 + \sqrt{60} -\sqrt{24} - \sqrt{40}} = \sqrt3 + \sqrt5 -\sqrt2$
b) Ta có:
$6 + \sqrt{24} + \sqrt{12} + \sqrt8$
$= 1 + 2 + 3 + 2\sqrt6 + 2\sqrt3 + 2\sqrt2$
$= 1^2 + (\sqrt2)^2 + (\sqrt3)^2 + 2\sqrt2.\sqrt3 + 2\sqrt3.1 + 2\sqrt2.1$
$= (1 + \sqrt2 + \sqrt3)^2$
Do đó:
$\sqrt{6 + \sqrt{24} + \sqrt{12} + \sqrt8} = 1 + \sqrt2 + \sqrt3$
Hay $\sqrt{6 + \sqrt{24} + \sqrt{12} + \sqrt8} -\sqrt3 = \sqrt2 + 1$
