# Bài 33 trang 19 SGK Toán 9 tập 1

##### Hướng dẫn giải

+ $$\sqrt{x^2}=|x|$$

+ $$|x|=x$$   nếu $$x \ge 0$$.

$$|x|=-x$$  nếu $$x<0$$.

+$$\dfrac{\sqrt a}{\sqrt b}=\sqrt{\dfrac{a}{b}}$$.

Lời giải chi tiết

a)

$$\sqrt{2}.x - \sqrt{50} = 0$$

$$\Leftrightarrow \sqrt{2}x=\sqrt{50}$$

$$\Leftrightarrow x=\dfrac{\sqrt{50}}{\sqrt{2}}$$

$$\Leftrightarrow x =\sqrt{\dfrac{50}{2}}$$

$$\Leftrightarrow x= \sqrt{25}$$

$$\Leftrightarrow x= \sqrt{5^2}$$

$$\Leftrightarrow x=5$$.

Vậy $$x=5$$.

b)

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

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

$$\Leftrightarrow \sqrt{3}.x=\sqrt{4.3}+\sqrt{9.3}- \sqrt{3}$$

$$\Leftrightarrow \sqrt{3}.x=\sqrt{4}. \sqrt{3}+\sqrt{9}. \sqrt{3}- \sqrt{3}$$

$$\Leftrightarrow \sqrt{3}.x=\sqrt{2^2}. \sqrt{3}+\sqrt{3^3}. \sqrt{3}- \sqrt{3}$$

$$\Leftrightarrow \sqrt{3}.x=2 \sqrt{3}+3\sqrt{3}- \sqrt{3}$$

$$\Leftrightarrow \sqrt{3}.x=(2+3-1).\sqrt{3}$$

$$\Leftrightarrow \sqrt{3}.x=4\sqrt{3}$$

$$\Leftrightarrow x=4$$.

Vậy $$x=4$$.

c)

$$\sqrt{3}x^2-\sqrt{12}=0$$

$$\Leftrightarrow \sqrt{3}x^2=\sqrt{12}$$

$$\Leftrightarrow \sqrt{3}x^2=\sqrt{4.3}$$

$$\Leftrightarrow \sqrt{3}x^2=\sqrt{4}.\sqrt 3$$

$$\Leftrightarrow x^2=\sqrt{4}$$

$$\Leftrightarrow x^2=\sqrt{2^2}$$

$$\Leftrightarrow x^2=2$$

$$\Leftrightarrow \sqrt{x^2}=\sqrt{2}$$

$$\Leftrightarrow |x|= \sqrt 2$$

$$\Leftrightarrow x= \pm \sqrt 2$$.

Vậy $$x= \pm\sqrt 2$$.

d)

$$\dfrac{x^{2}}{\sqrt{5}}- \sqrt{20} = 0$$

$$\Leftrightarrow \dfrac{x^2}{\sqrt{5}}=\sqrt{20}$$

$$\Leftrightarrow x^2=\sqrt{20}.\sqrt{5}$$

$$\Leftrightarrow x^2=\sqrt{20.5}$$

$$\Leftrightarrow x^2=\sqrt{100}$$

$$\Leftrightarrow x^2=\sqrt{10^2}$$

$$\Leftrightarrow x^2=10$$

$$\Leftrightarrow \sqrt{x^2}=\sqrt {10}$$

$$\Leftrightarrow |x|=\sqrt{10}$$

$$\Leftrightarrow x=\pm \sqrt{10}$$.

Vậy $$x= \pm \sqrt{10}$$.