Bài 33 (Trang 19 SGK Toán 9, Tập 1):

Giải phương trình:

$\mathrm{a}) \sqrt{2}.\mathrm{x} - \sqrt{50} = 0;$

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

$\mathrm{c}) \sqrt{3}.{\mathrm{x}}^2 - \sqrt{12} = 0;$

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

Hướng dẫn giải:

$$\begin{gathered} \mathrm{a}) \sqrt{2}.\mathrm{x} - \sqrt{50} = 0 \\ \iff \sqrt{2}\mathrm{x} = \sqrt{50} \\ \iff \mathrm{x} = \sqrt{\frac{50}{2}} \\ \iff \mathrm{x} = \sqrt{25} \\ \iff \mathrm{x} = 5 \end{gathered}$$

Vậy x = 5.

$$\begin{gathered} \mathrm{b}) \sqrt{3}.\mathrm{x} +\sqrt{3} = \sqrt{12} + \sqrt{27} \\ \iff \sqrt{3}x = \sqrt{12} + \sqrt{27} - \sqrt{3} \\ \iff \sqrt{3}x = \sqrt{4.3} + \sqrt{9.3} - \sqrt{3} \\ \iff \sqrt{3}x = 2\sqrt{3} + 3\sqrt{3} - \sqrt{3} \\ \iff x = 2 + 3 - 1 \\ \iff x = 4 \end{gathered}$$

Vậy x = 4.

$$\begin{gathered} \mathrm{c}) \sqrt{3}.{\mathrm{x}}^2 - \sqrt{12} = 0 \\ \iff \sqrt{3}{\mathrm{x}}^2 - \sqrt{12} = 0 \\ \iff \sqrt{3}{\mathrm{x}}^2 = \sqrt{4.3} \\ \iff \sqrt{3}{\mathrm{x}}^2 = \sqrt{4}.\sqrt{3} \\ \iff {\mathrm{x}}^2 = \sqrt{4} \\ \iff {\mathrm{x}}^2 = 2 \\ \iff \sqrt{{\mathrm{x}}^2} = \sqrt{2} \\ \iff \left|\mathrm{x}\right| = \sqrt{2} \\ \iff \mathrm{x} = \pm \sqrt{2} \\ \mathrm{Vậy} \mathrm{x} = \pm \sqrt{2} \end{gathered}$$