Bài 2:
a) Ta có: \(\dfrac{3-2\sqrt{3}}{\sqrt{3}}+\dfrac{6}{3+\sqrt{3}}\)
\(=\sqrt{3}-2+3-\sqrt{3}\)
=1
b) Ta có: \(\left(\dfrac{1}{2-\sqrt{5}}+\dfrac{2}{\sqrt{5}+\sqrt{3}}\right):\dfrac{1}{\sqrt{21-12\sqrt{3}}}\)
\(=\left(-\sqrt{5}-2+\sqrt{5}-\sqrt{3}\right):\dfrac{1}{2\sqrt{3}-3}\)
\(=-\left(2+\sqrt{3}\right)\cdot\sqrt{3}\left(2-\sqrt{3}\right)\)
\(=-\sqrt{3}\)
Bài 1:
a) Ta có: \(\sqrt{11+4\sqrt{6}}-\sqrt{11-4\sqrt{6}}\)
\(=3+2\sqrt{2}-3+2\sqrt{2}\)
\(=4\sqrt{2}\)
b) Ta có: \(\sqrt{14-8\sqrt{3}}-\sqrt{24-12\sqrt{3}}\)
\(=2\sqrt{2}-\sqrt{6}-3\sqrt{2}+\sqrt{6}\)
\(=-\sqrt{2}\)
a) \(\sqrt{11+4\sqrt{6}}-\sqrt{11-4\sqrt{6}}\)
=\(\sqrt{\left(\sqrt{2}\right)^2+\left(\sqrt{3}\right)^2+\left(\sqrt{2}\right)^2+2.\sqrt{2}.\sqrt{2}+2.\sqrt{3}.\sqrt{2}+2.\sqrt{3}.\sqrt{2}}\)
\(-\sqrt{\left(\sqrt{2}\right)^2+\left(\sqrt{3}\right)^2+\left(\sqrt{2}\right)^2+2.\sqrt{2}.\sqrt{2}-2.\sqrt{3}.\sqrt{2}-2.\sqrt{3}.\sqrt{2}}\)
=\(\sqrt{\left(\sqrt{2}+\sqrt{2}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{2}+\sqrt{2}-\sqrt{3}\right)^2}\)
=\(\left|\sqrt{2}+\sqrt{2}+\sqrt{3}\right|-\left|\sqrt{2}+\sqrt{2}-\sqrt{3}\right|=\sqrt{2}+\sqrt{2}+\sqrt{3}-\sqrt{2}-\sqrt{2}+\sqrt{3}\)
=\(2\sqrt{3}\)
B1.
b)\(\sqrt{14-8\sqrt{3}}-\sqrt{24-12\sqrt{3}}=\sqrt{2\left(7-4\sqrt{3}\right)}-\sqrt{2\left(12-6\sqrt{3}\right)}\)
=\(\sqrt{2\left(2^2-2.2.\sqrt{3}+\left(\sqrt{3}\right)^2\right)}-\sqrt{2\left(3^2-2.3.\sqrt{3}+\left(\sqrt{3}\right)^2\right)}\)
=\(\sqrt{2\left(2-\sqrt{3}\right)^2}-\sqrt{2\left(3-\sqrt{3}\right)^2}=\sqrt{2}\left(2-\sqrt{3}\right)-\sqrt{2}\left(3-\sqrt{3}\right)\)
=\(\sqrt{2}\left(2-\sqrt{3}-3+\sqrt{3}\right)=-\sqrt{2}\)
B2.
a) \(\dfrac{3-2\sqrt{3}}{\sqrt{3}}+\dfrac{6}{3+\sqrt{3}}=\dfrac{\sqrt{3}\left(\sqrt{3}-2\right)}{\sqrt{3}}+\dfrac{6\left(3-\sqrt{3}\right)}{9-3}\)
=\(\sqrt{3}-2+3-\sqrt{3}=1\)