(√3-3)/(√(2-√3) +2√2) +(√3+3)/(√(2+√3) -2√2)
$\dfrac{\sqrt{3}-3}{\sqrt{2-\sqrt{3}}+2\sqrt{2}}+\dfrac{\sqrt{3}+3}{\sqrt{2+\sqrt{3}}-2\sqrt{2}}
\\=\dfrac{\sqrt{2}(\sqrt 3-3)}{\sqrt{4-2\sqrt{3}}+4}+\dfrac{\sqrt{2}(\sqrt 3+3)}{\sqrt{4+2\sqrt{3}}-4}
\\=\dfrac{\sqrt{2}(\sqrt 3-3)}{\sqrt{(\sqrt{3}-1)^2}+4}+\dfrac{\sqrt{2}(\sqrt 3+3)}{\sqrt{(\sqrt{3}+1)^2}-4}
\\=\dfrac{\sqrt 2(\sqrt 3-3)}{\sqrt 3+3}+\dfrac{\sqrt 2(\sqrt 3+3)}{\sqrt 3-3}
\\=\dfrac{\sqrt 2[(\sqrt 3-3)^2+(\sqrt 3+3)^2]}{(\sqrt 3+3)(\sqrt 3-3)}
\\=\dfrac{\sqrt 2(3-6\sqrt 3+9+3+6\sqrt 3+9)}{3-9}
\\=\dfrac{24\sqrt 2}{-6}=-4\sqrt 2$
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
