D=\(\sqrt{4+2\sqrt{3}}-\sqrt{\dfrac{2}{2+\sqrt{3}}}\)
D=\(\sqrt{\left(\sqrt{3}+2\right)^2}-\sqrt{\dfrac{2.\left(2-\sqrt{3}\right)}{\left(2-\sqrt{3}\right).\left(2+\sqrt{3}\right)}}\)
D=\(\sqrt{3}+2-\sqrt{\dfrac{4-2\sqrt{3}}{4-3}}\)
D=\(\sqrt{3}+2-\dfrac{\sqrt{\left(\sqrt{3}-2\right)^2}}{1}\)
D=\(\sqrt{3}+2-\sqrt{3}+2\)
D=4
Ta có:
\(D=\sqrt{4+2\sqrt{3}}-\sqrt{\dfrac{2}{2+\sqrt{3}}}=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\dfrac{2\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}\)
\(=\sqrt{3}+1-\sqrt{\dfrac{4-2\sqrt{3}}{4-3}}=\sqrt{3}+1-\sqrt{\dfrac{\left(\sqrt{3}-1\right)^2}{1}}\)
\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)=\sqrt{3}+1-\sqrt{3}+1=2\)
