Bài tập sách giáo khoa: Biến đổi đơn giản biểu thức chứa căn thức bậc hai

a) $\sqrt{54}=\sqrt{9.6}=\sqrt{9}.\sqrt{6}=3\sqrt{6}$.

b) $\sqrt{108}=\sqrt{36.3}=\sqrt{36}.\sqrt{3}=6\sqrt{3}$.

c) $0,1\sqrt{20000}=0,1\sqrt{2.10000}=0,1 \sqrt{2}.\sqrt{10000}$

$=0,1.100\sqrt{2}=10\sqrt{2}$.

d) $-0,05\sqrt{28800}=-0,05\sqrt{288.100}$

$=-0,05\sqrt{2.144.100}=-0,05.\sqrt{2}.\sqrt{144}.\sqrt{100}$

$=-0,05.12.10\sqrt{2}$

$=-6 \sqrt{2}$.
e) $\sqrt{7.63.a^2}=\sqrt{7.7.9.a^2}=\sqrt{7^2}.\sqrt{9}.\sqrt{a^2}$

$=7.3.|a|=21|a|$

\(=\left[{}\begin{matrix}21a\text{ nếu }a\ge0\\-21a\text{ nếu }a< 0\end{matrix}\right.\).

a) $3\sqrt{5}=\sqrt{3^{2}.5}=\sqrt{9.5}=\sqrt{45}$.

b) $-5 \sqrt{2}=-\sqrt{5^{2}.2}=-\sqrt{25.2}=-\sqrt{50}$.

c) Vì $xy\ge 0$ do đó biểu thức $\sqrt{xy}$ có nghĩa.

Ta có: $-\dfrac{2}{3}\sqrt{xy}=-\sqrt{\left(\dfrac{2}{3}\right)^2.xy}=-\sqrt{\dfrac{4xy}{9}}$.
d) Với $x>0$ thì $\sqrt{\dfrac{2}{x}}$ có nghĩa.
$x\sqrt{\dfrac{2}{x}}=\sqrt{x^{2}\dfrac{2}{x}}=\sqrt{2x}$.

a) Ta có: $3 \sqrt{3}=\sqrt{3^2.3}=\sqrt{9.3}=\sqrt{27}$

Vì $\sqrt{27}>\sqrt{12}$ nên $3 \sqrt{3}>\sqrt{12}$

Vậy $3 \sqrt{3}>\sqrt{12}$.
b) Ta có: $3 \sqrt{5}=\sqrt{3^{2}. 5}=\sqrt{45}$

$7=\sqrt{7^{2}}=\sqrt{49}$

Vì $\sqrt{49}>\sqrt{45}$ nên $7>3 \sqrt{5}$

Vậy $7>3 \sqrt{5}$.

c) Ta có: $\dfrac{1}{3} \sqrt{51}=\sqrt{\left(\dfrac{1}{3}\right)^{2} .51}=\sqrt{\dfrac{51}{9}}$

$\dfrac{1}{5} \sqrt{150}=\sqrt{\left(\dfrac{1}{5}\right)^2.150}=\sqrt{\dfrac{150}{25}}=\sqrt{6}=\sqrt{\dfrac{6.9}{9}}=\sqrt{\dfrac{54}{9}}$

Vì $\sqrt{\dfrac{54}{9}}>\sqrt{\dfrac{51}{9}}$ nên $\dfrac{1}{3} \sqrt{51}<\dfrac{1}{5} \sqrt{150}$

Vậy $\dfrac{1}{3} \sqrt{51}<\dfrac{1}{5} \sqrt{150}$.

d) Ta có: $\dfrac{1}{2}\sqrt{6}=\sqrt{\left(\dfrac{1}{2}\right)^2.6}=\sqrt{\dfrac{6}{4}}$

$=\sqrt{\dfrac{3}{2}}=\sqrt{3.\dfrac{1}{2}}=\sqrt{3}.\sqrt{\dfrac{1}{2}}$

Vì $\sqrt{3}.\sqrt{\dfrac{1}{2}}<6 \sqrt{\dfrac{1}{2}}$ nên $\dfrac{1}{2}.\sqrt{6}<6\sqrt{\dfrac{1}{2}}$

Vậy $\dfrac{1}{2}\sqrt{6}<6\sqrt{\dfrac{1}{2}}$.

a) Với $x \geq 0$ thì $\sqrt{3 x}$ có nghĩa.

Ta có: $2 \sqrt{3x}-4 \sqrt{3 x}+27-3 \sqrt{3 x}$

$=2 \sqrt{3 x}-4 \sqrt{3 x}-3 \sqrt{3 x}+27$

$=-5 \sqrt{3 x}+27$.
b) Với $x \geq 0$ thì $\sqrt{2 x},\sqrt{8x},\sqrt{18x}$ có nghĩa.

Ta có: $3 \sqrt{2 x}-5 \sqrt{8 x}+7 \sqrt{18 x}+28$

$=3 \sqrt{2 x}-5 \sqrt{4. 2 x}+7 \sqrt{9.2 x}+28$

$=3 \sqrt{2 x}-5.\sqrt{4}.\sqrt{2 x}+7.\sqrt{9}.\sqrt{2 x}+28$

$=3 \sqrt{2 x}-5.2.\sqrt{2 x}+7.3.\sqrt{2 x}+28$

$=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+28$
$=14 \sqrt{2 x}+28=14(\sqrt{2 x}+2)$.

a) $\dfrac{2}{x^2-y^2}\sqrt{\dfrac{3(x+y)^2}{2}}$

$=\dfrac{2}{x^2-y^2}\sqrt{\dfrac{3}{2}.(x+y)^2}$

$=\dfrac{2}{x^2-y^2}\sqrt{(x+y)^2}.\sqrt{\dfrac{3}{2}}$

$=\dfrac{2}{x^2-y^2}.|x+y|\sqrt{\dfrac{3}{2}}$

$=\dfrac{|x+y|}{x^{2}-y^{2}}\sqrt{\dfrac{3.2^2}{2}}$

$=\dfrac{x+y}{(x-y)(x+y)}.\sqrt{6}$ (Do $x \ge 0, y \ge 0$ $\Rightarrow$ $x+y \ge 0$ nên $|x+y|=x+y$)

$=\dfrac{\sqrt{6}}{x-y}$.

b) $\dfrac{2}{2 a-1} \sqrt{5a^2(1-4 a+4 a^2)}$

$=\dfrac{2}{2 a-1} \sqrt{5a^2(1-2.1.2a+(2a)^2)}$

$=\dfrac{2}{2 a-1} \sqrt{5a^2(1-2a)^2}$

$=\dfrac{2}{2 a-1} \sqrt{5}.\sqrt{a^2}.\sqrt{(1-2a)^2}$

$=\dfrac{2}{2 a-1} \sqrt{5}.|a|.|1-2a|$

$=\dfrac{2}{2 a-1} \sqrt{5}.a.(2a-1)$ (Do $a>0,5$ $\Rightarrow$ $2a-1>0$ nên $|a|=a$ và $|1-2a|=2a-1$)

$=2a\sqrt{5}$.