Áp dụng quy tắc chia hai căn bậc hai, hãy tính :

a) 2300−−−−√23−−√ = \(\sqrt{\dfrac{2300}{23}}\) = \(\sqrt{100}\) = 10

b) 12,5−−−−√0,5−−−√ = \(\sqrt{\dfrac{12,5}{0,5}}\) = \(\sqrt{25}\) = 5

c) 192−−−√12−−√ = \(\sqrt{\dfrac{192}{12}}\) = \(\sqrt{16}\) = 4

d) 6–√150−−−√ = \(\sqrt{\dfrac{6}{150}}\) = \(\sqrt{\dfrac{1}{25}}\) = \(\dfrac{1}{5}\)

12 , 5 0 , 5 = 12 , 5 0 , 5 = 25 = 5

6 150 = 6 150 = 1 25 = 1 5

192 12 = 192 12 = 16 = 4

\(3,\dfrac{\sqrt{9+6\sqrt{2}}}{\sqrt{3}}\)

\(=\dfrac{\sqrt{3\left(3+2\sqrt{2}\right)}}{\sqrt{3}}\)

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

\(=\sqrt{3+2\sqrt{2}}\)

\(=\sqrt{\sqrt{2^2}+2\sqrt{2}+1}\)

\(=\sqrt{\left(\sqrt{2}+1\right)^2}\)

\(=\left|\sqrt{2}+1\right|=\sqrt{2}+1\)

\(4,\sqrt{2+\sqrt{3}}:\sqrt{\dfrac{1}{2}}\)

\(=\sqrt{2+\sqrt{3}}:\dfrac{1}{\sqrt{2}}\)

\(=\sqrt{2+\sqrt{3}}.\sqrt{2}\)

\(=\sqrt{4+2\sqrt{3}}\)

\(=\sqrt{\sqrt{3^2}+2\sqrt{3}+1}\)

\(=\sqrt{\left(\sqrt{3}+1\right)}^2\)

\(=\left|\sqrt{3}+1\right|=\sqrt{3}+1\)