Ta có :

a.  $\sqrt{2}x-\sqrt{50}=0$ 

<=> $\sqrt{2}x-\sqrt{25}.\sqrt{2}=0$

<=> $\sqrt{2}x-5.\sqrt{2}=0$

<=> $\sqrt{2}x=5.\sqrt{2}$

<=> $x=5$

Vậy $x=5$ .

b.  $\sqrt{3}x+\sqrt{3}=\sqrt{12}+\sqrt{27}$

<=> $\sqrt{3}x+\sqrt{3}=\sqrt{3}.\sqrt{4}+\sqrt{9}.\sqrt{3}$

<=> $\sqrt{3}x=2\sqrt{3}+3\sqrt{3}-\sqrt{3}$

<=> $\sqrt{3}x=4\sqrt{3}$

<=> $x=4$

Vậy $x=4$ .

c.  $\sqrt{3}x^{2}-\sqrt{12}=0$

<=> $\sqrt{3}x^{2}=\sqrt{4}.\sqrt{3}$

<=> $\sqrt{3}x^{2}=2\sqrt{3}$

<=> $x^{2}=2<=>x=\pm \sqrt{2}$

Vậy $x=\pm \sqrt{2}$

d.  $\frac{x^{2}}{\sqrt{5}}-\sqrt{20}=0$

<=> $\frac{x^{2}}{\sqrt{5}}=\sqrt{20}$

<=> $x^{2}=\sqrt{20}.\sqrt{5}$

<=> $x^{2}=10<=>x=\pm \sqrt{10}$

Vậy $x=\pm \sqrt{10}$