a) $\sqrt{3}x$ `=` $\sqrt{27}$
⇔ $\sqrt{3}x$ `=` $3\sqrt{3}$
⇔ `x =3`
Vậy `x = 3`
b) $\sqrt{3}x$ `-` $\sqrt{27}$ = $\sqrt{12}$ `-` $\sqrt{75}$
⇔ $\sqrt{3}x$ `=` $-3\sqrt{3}$ `+` $2\sqrt{3}x$ `-` $5\sqrt{3}$
⇔ $\sqrt{3}x$ `=` $-6\sqrt{3}$
⇔ `x = -6`
Vậy `x = -6`
c) $\sqrt{5}x^2$ `-` $\sqrt{20}$ `= 0`
⇔ $\sqrt{5}x^2$ `=` $2\sqrt{5}$
⇔ `x^2 = 2`
⇔ \(\left[ \begin{array}{l}x=\sqrt{2}\\x=-\sqrt{2}\end{array} \right.\)
Vậy `x=\sqrt{2}` hoặc `x=-\sqrt{2}`
d) $\dfrac{2x^2}{\sqrt{3}}$ `-` $\sqrt{12}$ `= 0`
⇔ $\dfrac{2x^2}{\sqrt{3}}$ `=` $2\sqrt{3}$
⇔ $2\sqrt{3}x^2$ `=` `6`
⇔ $\sqrt{3}x^2$ `= 3`
⇔ `x^2` = $\sqrt{3}$
⇔ \(\left[ \begin{array}{l}x=\sqrt{\sqrt{3}}\\x=-\sqrt{\sqrt{3}}\end{array} \right.\)
Vậy `x=\sqrt{\sqrt{3}}` hoặc `x=-\sqrt{\sqrt{3}}`
