b, \(\dfrac{2}{\sqrt{5}+2}+\dfrac{2}{2-\sqrt{5}}\)
\(=\dfrac{2\left(\sqrt{5}-2\right)}{5-4}-\dfrac{2\left(\sqrt{5}+2\right)}{5-4}\)
\(=2\sqrt{5}-4-2\sqrt{5}-4=-8\)
a, \(\sqrt{2}\left(\sqrt{8}+\sqrt{32}-\sqrt{98}\right)\)
\(=\sqrt{2}\left(2\sqrt{2}+4\sqrt{2}-7\sqrt{2}\right)\)
\(=\sqrt{2}.\left(-\sqrt{2}\right)=-2\)
c, \(\left(2+\sqrt{3}\right)\sqrt{11-6\sqrt{2}}\)
\(=\left(2+\sqrt{3}\right)\sqrt{\left(\sqrt{2}-3\right)^2}\)
\(=\left(2+\sqrt{3}\right)\left(3-\sqrt{2}\right)\)
\(=6-2\sqrt{2}+3\sqrt{3}-\sqrt{6}\)
a: Ta có: \(\sqrt{2}\left(\sqrt{8}+\sqrt{32}-\sqrt{98}\right)\)
\(=\sqrt{2}\left(2\sqrt{2}+4\sqrt{2}-7\sqrt{2}\right)\)
=-2
b: Ta có: \(\dfrac{2}{\sqrt{5}+2}-\dfrac{2}{\sqrt{5}-2}\)
\(=2\sqrt{5}-4-2\sqrt{5}-4\)
=-8
c: Ta có: \(\left(2+\sqrt{3}\right)\cdot\sqrt{11-6\sqrt{2}}\)
\(=\left(2+\sqrt{3}\right)\left(3-\sqrt{2}\right)\)
\(=6-2\sqrt{2}+3\sqrt{3}-\sqrt{6}\)