$a)\dfrac{\sqrt{10}-\sqrt{15}}{\sqrt{8}-\sqrt{12}}$
$=\dfrac{\sqrt{5}(\sqrt{2}-\sqrt{3})}{\sqrt{4}(\sqrt{2}-\sqrt{3})}$
$=\dfrac{\sqrt{5}}{\sqrt{4}}$
$b)\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}$
$=\dfrac{\sqrt{5}(\sqrt{3}-1)}{\sqrt{3}-1}+\dfrac{\sqrt{5}(\sqrt{5}-2)}{2(\sqrt{5}-2)}$
$=\sqrt{5}+\dfrac{\sqrt{5}}{2}$
$=\dfrac{3\sqrt{5}}{2}$
Bài `5B:`
$a)\dfrac{\sqrt{6}-\sqrt{15}}{\sqrt{35}-\sqrt{14}}$
$=\dfrac{\sqrt{3}(\sqrt{2}-\sqrt{5})}{\sqrt{7}(\sqrt{5}-\sqrt{2})}$
$=\dfrac{-\sqrt{3}}{\sqrt{7}}$
$b)\dfrac{5+\sqrt{5}}{\sqrt{10}+\sqrt{2}}$
$=\dfrac{\sqrt{5}(\sqrt{5}+1)}{\sqrt{2}(\sqrt{5}+1)}$
$=\dfrac{\sqrt{5}}{\sqrt{2}}$
Bài `6B:`
$a)\sqrt{\dfrac{-2t}{3}}.\sqrt{\dfrac{-3t}{8}}$ `(x<=0)`
$=\sqrt{\dfrac{-2t.-3t}{3.8}}$
$=\sqrt{\dfrac{t^2}{4}}$
$=\dfrac{|t|}{2}$
$=\dfrac{-t}{2}$
$b)\sqrt{x-\sqrt{x^2-1}}.\sqrt{x+\sqrt{x^2-1}}$ `(x>=1)`
$=\sqrt{(x-\sqrt{x^2-1})(\sqrt{x}+\sqrt{x^2-1})}$
$=\sqrt{x^2-(\sqrt{x^2-1})^2}$
$=\sqrt{x^2-x^2+1}$
$=\sqrt{1}=1$
