a) ta có : \(\sqrt{8+4\sqrt{3}}-\sqrt{8-4\sqrt{3}}=\dfrac{\sqrt{16+8\sqrt{3}}-\sqrt{16-8\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{\left(2\sqrt{3}+2\right)^2}-\sqrt{\left(2\sqrt{3}-2\right)^2}}{\sqrt{2}}=\dfrac{2\sqrt{3}+2-2\sqrt{3}+2}{\sqrt{2}}\)
\(=\dfrac{4}{\sqrt{2}}=2\sqrt{2}\)
b) ta có : \(\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}+2\right)^2}\)
\(=\sqrt{5}-2-\sqrt{5}-2=-4\)
a/ Đặt: \(A=\sqrt{8+4\sqrt{3}}-\sqrt{8-4\sqrt{3}}\)
\(\Leftrightarrow A^2=8+4\sqrt{3}-2\sqrt{\left(8+4\sqrt{3}\right)\left(8-4\sqrt{3}\right)}+8-4\sqrt{3}=16-2\sqrt{16}=16-2\cdot4=8\)
\(\Leftrightarrow A=\sqrt{8}=2\sqrt{2}\)
b/ Đặt:
\(B=\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}\)
\(\Leftrightarrow B^2=9-4\sqrt{5}-2\sqrt{\left(9-4\sqrt{5}\right)\left(9+4\sqrt{5}\right)}+9+4\sqrt{5}=18-2\sqrt{1}=18-2=16\)
mà ta thấy: \(\sqrt{9-4\sqrt{5}}< \sqrt{9+4\sqrt{5}}\Rightarrow B< 0\Rightarrow B=-\sqrt{16}=-4\)
