\(\sqrt{\dfrac{3\sqrt{3}-4}{2\sqrt{3}+1}}-\sqrt{\dfrac{\sqrt{3}+4}{5-2\sqrt{3}}}=\sqrt{\dfrac{\left(3\sqrt{3}-4\right)\left(2\sqrt{3}-1\right)}{\left(2\sqrt{3}+1\right)\left(2\sqrt{3}-1\right)}}-\sqrt{\dfrac{\left(\sqrt{3}+4\right)\left(5+2\sqrt{3}\right)}{\left(5-2\sqrt{3}\right)\left(5+2\sqrt{3}\right)}}=\sqrt{\dfrac{18-3\sqrt{3}-8\sqrt{3}+4}{\left(2\sqrt{3}\right)^2-1}}-\sqrt{\dfrac{5\sqrt{3}+6+20+8\sqrt{3}}{25-\left(2\sqrt{3}\right)^2}}=\sqrt{\dfrac{22-11\sqrt{3}}{11}}-\sqrt{\dfrac{26+13\sqrt{3}}{13}}=\sqrt{\dfrac{11\left(2-\sqrt{3}\right)}{11}}-\sqrt{\dfrac{13\left(2+\sqrt{3}\right)}{13}}=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)Đặt \(A=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\Rightarrow A^2=2-\sqrt{3}+2+\sqrt{3}-2\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=4-2\sqrt{4-3}=4-2\sqrt{1}=2\Rightarrow A=\pm\sqrt{2}\)Ta có \(\left(2-\sqrt{3}\right)< \left(2+\sqrt{3}\right)\Rightarrow\sqrt{2-\sqrt{3}}< \sqrt{2+\sqrt{3}}\Rightarrow\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}< 0\Rightarrow A< 0\)Vậy A=\(-\sqrt{2}\)
