Bài 48 trang 29 SGK Toán 9 tập 1 — Không quảng cáo

Đề bài

Khử mẫu của biểu thức lấy căn

$\sqrt{\dfrac{1}{600}};\,\,\sqrt{\dfrac{11}{540}};\,\,\sqrt{\dfrac{3}{50}};\,\,\sqrt{\dfrac{5}{98}}; \,\,\sqrt{\dfrac{(1-\sqrt{3})^{2}}{27}}.$

Lời giải chi tiết

+$\sqrt{\dfrac{1}{600}}=\dfrac{\sqrt 1}{\sqrt{600}}$$=\dfrac{ 1}{\sqrt{6.100}}$$=\dfrac{1}{\sqrt{6.10^2}}$

$=\dfrac{ 1}{\sqrt{6}.\sqrt{10^2}}$$=\dfrac{ 1}{10\sqrt{6}}$$=\dfrac{ 1.\sqrt 6}{10.6}$$=\dfrac{ \sqrt 6}{60}$

+$\sqrt{\dfrac{11}{540}}=\dfrac{\sqrt{11}}{\sqrt{540}}=\dfrac{\sqrt{11}}{\sqrt{36.15}}$

$=\dfrac{\sqrt{11}}{\sqrt{36}.\sqrt{15}}=\dfrac{\sqrt{11}}{\sqrt{6^2}.\sqrt{15}}$

$=\dfrac{\sqrt{11}}{6\sqrt{15}}=\dfrac{\sqrt{11}.\sqrt{15}}{6.15}$

$=\dfrac{\sqrt{11.15}}{90}=\dfrac{\sqrt{165}}{90}$.

+ $\sqrt{\dfrac{3}{50}}=\dfrac{\sqrt 3}{\sqrt{50}}=\dfrac{\sqrt 3}{\sqrt{25.2}}=\dfrac{\sqrt{3}}{\sqrt{25}.\sqrt{2}}$

$=\dfrac{\sqrt{3}}{\sqrt{5^2}.\sqrt{2}}=\dfrac{\sqrt{3}}{5\sqrt{2}}=\dfrac{\sqrt{3}.\sqrt 2}{5.2}$

$=\dfrac{\sqrt{3.2}}{10}=\dfrac{\sqrt{6}}{10}$

+ $\sqrt{\dfrac{5}{98}}=\dfrac{\sqrt 5}{\sqrt {98}}=\dfrac{\sqrt 5}{\sqrt{49.2}}=\dfrac{\sqrt 5}{\sqrt{49}\sqrt{2}}$

$=\dfrac{\sqrt 5}{\sqrt{7^2}.\sqrt 2}=\dfrac{\sqrt 5}{7\sqrt 2}=\dfrac{\sqrt 5 . \sqrt 2}{7. 2}$

$=\dfrac{\sqrt {5. 2}}{14}=\dfrac{\sqrt{10}}{14}$.

+$\sqrt{\dfrac{(1-\sqrt{3})^{2}}{27}}=\dfrac{\sqrt{(1-\sqrt 3)^2}}{\sqrt {27}}=\dfrac{\sqrt{(1-\sqrt 3)^2}}{\sqrt {9.3}}$

$=\dfrac{\sqrt{(1-\sqrt 3)^2}}{\sqrt {3^2.3}}$$=\dfrac{|1-\sqrt{3}|}{3\sqrt {3}}$

Vì $1< 3 \Leftrightarrow \sqrt 1 < \sqrt 3 \Leftrightarrow 1< \sqrt 3$ $\Leftrightarrow 1- \sqrt 3 < 0$

$\Leftrightarrow |1- \sqrt 3|=-(1-\sqrt 3)=-1 + \sqrt 3 = \sqrt 3 -1.$

Do đó: $\dfrac{|1-\sqrt{3}|}{3\sqrt {3}}=\dfrac{\sqrt{3}-1}{3\sqrt {3}}=\dfrac{\sqrt 3(\sqrt{3}-1)}{9}=\dfrac{3-\sqrt 3}{9}.$