$a)\sqrt{4x-12}-2\sqrt{\dfrac{x-3}{4}}=2$ `(x≥3)`
$⇔2\sqrt{x-3}-\dfrac{2\sqrt{x-3}}{2}=2$
$⇔2\sqrt{x-3}-\sqrt{x-3}=2$
$⇔\sqrt{x-3}=2$
$⇔x-3=4$
$⇔x=7$
$b)\sqrt{9(5x-2)}+\sqrt{20x-8}=20$ `(x≥2/5)`
$⇔3\sqrt{5x-2}+2\sqrt{5x-2}=20$
$⇔5\sqrt{5x-2}=20$
$⇔\sqrt{5x-2}=4$
$⇔5x-2=16$
$⇔5x=17$
$⇔x=\dfrac{17}{5}$
$c)\sqrt{x^2-3(2x-3)}=2x-7$ `(x∈Z)`
$⇔\sqrt{x^2-6x+9}=2x-7$
$⇔\sqrt{(x-3)^2}=2x-7$
$⇔x-3=2x-7$
$⇔x=4$
Bài 2:
$a)\dfrac{8}{3\sqrt{5}}$
$=\dfrac{8\sqrt{5}}{15}$
$b)\sqrt{\dfrac{4}{2}}$
$=\dfrac{2\sqrt{2}}{2}$
$c)\dfrac{1+\sqrt{2}}{3\sqrt{2}}$
$=\dfrac{\sqrt{2}+2}{6}$
$d)\dfrac{1}{\sqrt{5}-2}$
$=\dfrac{\sqrt{5}+2}{(\sqrt{5}-2)(\sqrt{5}+2)}$
$=\dfrac{\sqrt{5}+2}{1}$
$=\sqrt{5}+2$
$e)\dfrac{4}{\sqrt{5}+1}$
$=\dfrac{4\sqrt{5}-4}{(\sqrt{5}+1)(\sqrt{5}-1)}$
$=\dfrac{4(\sqrt{5}-1)}{4}$
$=\sqrt{5}-1$
$f)\dfrac{2}{\sqrt{10}+\sqrt{6}}$
$=\dfrac{2\sqrt{10}-2\sqrt{6}}{(\sqrt{10}+\sqrt{6})(\sqrt{10}-\sqrt{6})}$
$=\dfrac{2(\sqrt{10}-\sqrt{6})}{4}$
$=\dfrac{\sqrt{10}-\sqrt{6}}{2}$
