Bài 48 trang 29 SGK Toán 9 tập 1

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