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

Hướng dẫn giải:

+) $\sqrt{\dfrac{1}{600}}=\dfrac{\sqrt{1}}{\sqrt{600}}=\dfrac{1}{\sqrt{6.10^{2}}}=\dfrac{1}{\sqrt{6}.\sqrt{10^2}}=\dfrac{1}{10\sqrt{6}}=\dfrac{1 \cdot \sqrt{6}}{10 \sqrt{6} \cdot \sqrt{6}}=\dfrac{\sqrt{6}}{60}$.

+) $\sqrt{\dfrac{11}{540}}=\dfrac{\sqrt{11}}{\sqrt{540}}=\dfrac{\sqrt{11}}{\sqrt{36.15}}=\dfrac{\sqrt{11}}{\sqrt{36}.\sqrt{15}}=\dfrac{\sqrt{11}}{6\sqrt{15}}$

$=\dfrac{\sqrt{11} \cdot \sqrt{15}}{6 \sqrt{15} \cdot \sqrt{15}}=\dfrac{\sqrt{11.15}}{6.15}=\dfrac{\sqrt{165}}{90}$.

+) $\sqrt{\dfrac{3}{50}}=\dfrac{\sqrt{3}}{\sqrt{50}}=\dfrac{\sqrt{3}}{\sqrt{25.2}}=\dfrac{\sqrt{3}}{\sqrt{25}\sqrt{2}}=\dfrac{\sqrt{3}}{5\sqrt{2}}=\dfrac{\sqrt{3} \cdot \sqrt{2}}{5 \sqrt{2} \cdot \sqrt{2}}=\dfrac{\sqrt{6}}{10}$.

+) $\sqrt{\dfrac{5}{98}}=\dfrac{\sqrt{5}}{\sqrt{98}}=\dfrac{\sqrt{5}}{\sqrt{49.2}}=\dfrac{\sqrt{5}}{\sqrt{49}.\sqrt{2}}=\dfrac{\sqrt{5}}{7\sqrt{2}}=\dfrac{\sqrt{5} \cdot \sqrt{2}}{7 \sqrt{2} \cdot \sqrt{2}}=\dfrac{\sqrt{10}}{14}$.

+) $\sqrt{\dfrac{(1-\sqrt{3})^{2}}{27}}=\dfrac{\sqrt{(1-\sqrt{3})^{2}}}{\sqrt{27}}=\dfrac{|1-\sqrt{3}|}{\sqrt{9.3}}=\dfrac{1-\sqrt{3}}{\sqrt{9}.\sqrt{3}}$(vì $\sqrt{3}>1$ nên $|1-\sqrt{3}|=\sqrt{3}-1$)

$=\dfrac{(\sqrt{3}-1) \cdot \sqrt{3}}{3 \sqrt{3} \cdot \sqrt{3}}=\dfrac{(\sqrt{3}-1) \cdot \sqrt{3}}{9}$

Hướng dẫn giải:

+) $\mathrm{ab} \sqrt{\dfrac{\mathrm{a}}{\mathrm{b}}}=\dfrac{\mathrm{ab}}{|\mathrm{b}|} \sqrt{\mathrm{ab}}$

+) $\dfrac{\mathrm{a}}{\mathrm{b}} \sqrt{\dfrac{\mathrm{b}}{\mathrm{a}}}=\dfrac{\mathrm{a}}{|\mathrm{a}| \cdot \mathrm{b}} \sqrt{\mathrm{ab}}$

Thu gọn kết quả hơn ta được:

-) Nếu $\mathrm{a}>0, \mathrm{~b}>0$, ta có:

$\dfrac{\mathrm{a}}{\mathrm{b}} \sqrt{\dfrac{\mathrm{b}}{\mathrm{a}}}=\dfrac{\sqrt{\mathrm{ab}}}{\mathrm{b}}$

-) Nếu $a<0, b<0$, ta có:

$\dfrac{\mathrm{a}}{\mathrm{b}} \sqrt{\dfrac{\mathrm{b}}{\mathrm{a}}}=\dfrac{-\sqrt{a b}}{b}$ 

+) $\sqrt{\dfrac{1}{\mathrm{~b}}+\dfrac{1}{\mathrm{~b}^{2}}}=\dfrac{\sqrt{\mathrm{b}+1}}{|\mathrm{~b}|}$

+) $\sqrt{\dfrac{9 \mathrm{a}^{3}}{36 \mathrm{~b}}}=\dfrac{\sqrt{\mathrm{a}^{3} \mathrm{~b}}}{2|\mathrm{~b}|}=\dfrac{|\mathrm{a}| \sqrt{\mathrm{ab}}}{2|\mathrm{~b}|}=\dfrac{\mathrm{a} \sqrt{\mathrm{ab}}}{2 \mathrm{~b}}$
+) $3 \mathrm{xy} \sqrt{\dfrac{2}{\mathrm{xy}}}=3 \mathrm{xy} \cdot \dfrac{\sqrt{2 \mathrm{xy}}}{\mathrm{xy}}=3 \sqrt{2 \mathrm{xy}}$

Có thể nhận xét $\mathrm{xy}>0$, dùng cách đưa } nhân tử $xy$ vào trong căn thức cũng được kết quả.

Hướng dẫn giải:

+) $\dfrac{5}{\sqrt{10}} =\dfrac{5 \cdot \sqrt{10}}{\sqrt{10} \cdot \sqrt{10}}=\dfrac{5 \cdot \sqrt{10}}{10}=\dfrac{\sqrt{10}}{2}$.

+) $\dfrac{1}{3 \sqrt{20}}=\dfrac{1}{3 \sqrt{2^{2} .5}}=\dfrac{1}{3.2 \sqrt{5}}=\dfrac{1 . \sqrt{5}}{6 \sqrt{5} \cdot \sqrt{5}}$

$=\dfrac{\sqrt{5}}{6.5}=\dfrac{\sqrt{5}}{30}$.

+) $\dfrac{5}{2 \sqrt{5}}=\dfrac{5 \sqrt{5}}{2 \cdot \sqrt{5} \cdot \sqrt{5}}=\dfrac{5 \cdot \sqrt{5}}{2.5}=\dfrac{\sqrt{5}}{2}$.

+) $\dfrac{2 \sqrt{2}+2}{5 \sqrt{2}}=\dfrac{(2 \sqrt{2}+2) \sqrt{2}}{5 \sqrt{2} \cdot \sqrt{2}}=\dfrac{2(\sqrt{2})^{2}+2 \sqrt{2}}{5.2}$

$=\dfrac{4+2 \sqrt{2}}{10}=\dfrac{2+\sqrt{2}}{5}$.

+) $\dfrac{y+b \sqrt{y}}{b.\sqrt{y}}=\dfrac{(y+b.\sqrt{y}) \sqrt{y}}{b.\sqrt{y} \cdot \sqrt{y}}=\dfrac{y \sqrt{y}+b y}{b y}=\dfrac{\sqrt{y}+b}{b}$.

Hướng dẫn giải:

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

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

+) $\dfrac{2+\sqrt{3}}{2-\sqrt{3}}=\dfrac{(2+\sqrt{3})(2+\sqrt{3)}}{(2-\sqrt{3})(2+\sqrt{3})}=\dfrac{4+4 \sqrt{3}+3}{2^{2}-(\sqrt{3})^{2}}=7+4 \sqrt{3}$.

+) $\dfrac{b}{3+\sqrt{b}}=\dfrac{b(3-\sqrt{b})}{(3+\sqrt{b})(3-\sqrt{b})}=\dfrac{b(3-\sqrt{b})}{3^{2}-(\sqrt{b})^{2}}=\dfrac{b(3-\sqrt{b})}{9-b}$.

+) $\dfrac{p}{2 .\sqrt{p}-1}=\dfrac{p(2 .\sqrt{p}+1)}{(2 .\sqrt{p}-1)(2 .\sqrt{p}+1)}=\dfrac{p(2 \sqrt{p}+1)}{(2 \sqrt{p})^{2}-1^{2}}=\dfrac{2 p.\sqrt{p}+p}{4 p-1}$.

Hướng dẫn giải:

+) $\dfrac{2}{\sqrt{6}-\sqrt{5}}=\dfrac{2(\sqrt{6}+\sqrt{5})}{(\sqrt{6}-\sqrt{5})(\sqrt{6}+\sqrt{5})}$

$=\dfrac{2(\sqrt{6}+\sqrt{5})}{6-5}=2(\sqrt{6}+\sqrt{5})$.
+) $\dfrac{3}{\sqrt{10}+\sqrt{7}}=\dfrac{3(\sqrt{10}-\sqrt{7})}{(\sqrt{10}+\sqrt{7})(\sqrt{10}-\sqrt{7})}$

$=\dfrac{3(\sqrt{10}-\sqrt{7})}{10-7}=\sqrt{10}-\sqrt{7}$.

+) $\dfrac{1}{\sqrt{x}-\sqrt{y}}=\dfrac{1(\sqrt{x}+\sqrt{y})}{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}$ (Do $\sqrt{x}+\sqrt{y} \ne 0$)

$=\dfrac{\sqrt{x}+\sqrt{y}}{x-y}$(do $x \neq y$ nên $\sqrt{x} \neq \sqrt{y}$).
+) $\dfrac{2 a b}{\sqrt{a}-\sqrt{b}}=\dfrac{2 a b(\sqrt{a}+\sqrt{b})}{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}$ (Do $\sqrt{a}+\sqrt{b} \ne 0$)

$=\dfrac{2 a b(\sqrt{a}+\sqrt{b})}{a-b}$(do $a \neq b$ nên $\sqrt{a} \neq \sqrt{b}$).

Hướng dẫn giải:

a) $2\sqrt{3}.(\sqrt{3}-\sqrt{2})=6-2\sqrt{6}.$

b) $\dfrac{\mathrm{ab}}{|\mathrm{ab}|} \sqrt{1+\mathrm{a}^{2} \mathrm{~b}^{2}}$. Rút gọn hơn, ta có kết quả

+) $\mathrm{ab}>0$ thì $a b \sqrt{1+\dfrac{1}{a^{2} b^{2}}}=\sqrt{1+\mathrm{a}^{2} \mathrm{~b}^{2}}$.

+) $\mathrm{ab}<0$ thì $a b \sqrt{1+\dfrac{1}{a^{2} b^{2}}}=-\sqrt{1+\mathrm{a}^{2} \mathrm{~b}^{2}}$.
c) $\dfrac{1}{b^{2}} \sqrt{a b+a}$.
d) Cách 1.

$\dfrac{\mathrm{a}+\sqrt{\mathrm{ab}}}{\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}}=\dfrac{(\mathrm{a}+\sqrt{\mathrm{ab}})(\sqrt{\mathrm{a}}-\sqrt{\mathrm{b}})}{(\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}})(\sqrt{\mathrm{a}}-\sqrt{\mathrm{b}})}$.

$=\dfrac{\mathrm{a} \sqrt{\mathrm{a}}+\sqrt{\mathrm{a}^{2} \mathrm{~b}}-\mathrm{a} \sqrt{\mathrm{b}}-\sqrt{\mathrm{ab}^{2}}}{\mathrm{a}-\mathrm{b}}=\frac{\sqrt{\mathrm{a}}(\mathrm{a}-\mathrm{b})}{\mathrm{a}-\mathrm{b}}=\sqrt{\mathrm{a}}$ . 
Cách 2.

$\dfrac{\mathrm{a}+\sqrt{\mathrm{ab}}}{\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}}=\dfrac{\sqrt{\mathrm{a}}(\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}})}{\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}}=\sqrt{\mathrm{a}}$ .

Hướng dẫn giải:

Gợi ý: Phân tích tử thành nhân tử để rút gọn nhân tử đó với mẫu. 

+) $\dfrac{2+\sqrt{2}}{1+\sqrt{2}}=\dfrac{(\sqrt{2})^{2}+\sqrt{2}}{1+\sqrt{2}}=\dfrac{\sqrt{2}(\sqrt{2}+1)}{1+\sqrt{2}}=\sqrt{2}$.

+) $\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}=\dfrac{\sqrt{3.5}-\sqrt{5}}{1-\sqrt{3}}=\dfrac{\sqrt{5}(\sqrt{3}-1)}{1-\sqrt{3}}=-\sqrt{5}$.

+) $\dfrac{2 \sqrt{3}-\sqrt{6}}{\sqrt{8}-2}=\dfrac{\sqrt{2^{2} \cdot 3}-\sqrt{6}}{\sqrt{4.2}-2}=\dfrac{\sqrt{2.6}-\sqrt{6}}{2 \sqrt{2}-2}=\dfrac{\sqrt{6}(\sqrt{2}-1)}{2(\sqrt{2}-1)}=\dfrac{\sqrt{6}}{2}$.

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

+) $\dfrac{p-2 \sqrt{p}}{\sqrt{p}-2}=\dfrac{(\sqrt{p})^{2}-2 \sqrt{p}}{\sqrt{p}-2}=\dfrac{\sqrt{p}(\sqrt{p}-2)}{\sqrt{p}-2}=\sqrt{p}$.

Hướng dẫn giải:

a) $(\sqrt{a}+1)(b \sqrt{a}+1)$.
b) $(x-y)(\sqrt{x}+\sqrt{y})$.

Hướng dẫn giải:

Biến đổi đưa thừa số vào trong dấu căn để so sánh (số nào lấy căn lớn hơn thì căn của số đó lớn hơn). Kết quả sắp xếp theo thứ tự tăng dần :
a) $2 \sqrt{6}, \sqrt{29}, 4 \sqrt{2}, 3 \sqrt{5} ;$

b) $\sqrt{38}, 2 \sqrt{14}, 3 \sqrt{7}, 6 \sqrt{2}$.

Hướng dẫn giải:

Chọn D.

$\sqrt{25 x}-\sqrt{16x}=9$

$\Leftrightarrow \sqrt{5^{2}.x}-\sqrt{4^{2}.x}=9$

$\Leftrightarrow 5 \sqrt{x}-4 \sqrt{x}=9$

$\Leftrightarrow \sqrt{x}=9$

$\Leftrightarrow x=81$.