Câu 1
a) $A = 2\sqrt{3} - \sqrt{12} - \sqrt{9} = 2\sqrt{3} - \sqrt{4.3} - 3 = 2\sqrt{3} - 2\sqrt{3} - 3 = -3$
b) $B = \sqrt{3} .\sqrt{12} + \sqrt{3} . \sqrt{27} = \sqrt{36} + \sqrt{81} = 6 + 9 = 15$.
c) $C = \sqrt{4 - 4\sqrt{3} + 3} + \sqrt{3 + 2\sqrt{3} + 1}$
$= \sqrt{2^2 - 2.2.\sqrt{3} + (\sqrt{3})^2} + \sqrt{(\sqrt{3})^2 + 2.1.\sqrt{3} + 1}$
$= \sqrt{(2-\sqrt{3})^2} + \sqrt{(\sqrt{3} + 1)^2}$
$= 2 - \sqrt{3} + \sqrt{3} + 1$
$= 2 +1 = 3$
Câu 2
Ta có
$(\dfrac{\sqrt{x}}{\sqrt{y}} - \dfrac{\sqrt{y}}{\sqrt{x}}) \dfrac{\sqrt{xy}}{\sqrt{x} - \sqrt{y}}$
$= \dfrac{x - y}{\sqrt{xy}} . \dfrac{\sqrt{xy}}{\sqrt{x} - \sqrt{y}}$
$= (\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y}) . \dfrac{1}{\sqrt{x} - \sqrt{y}}$
$= \sqrt{x} + \sqrt{y}$.
