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giúp mình với ạ >.< chi tiết nhá
$3)\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}=\frac{2\sqrt{5}(\sqrt{5}+\sqrt{2})}{\sqrt{5}+\sqrt{2}}=2\sqrt{5}
\\6)\sqrt{\dfrac{2-\sqrt 3}{2+\sqrt 3}}+\sqrt{\dfrac{2+\sqrt 3}{2-\sqrt 3}}=\sqrt{(2-\sqrt 3)^2}+\sqrt{(2+\sqrt 3)^2}=2-\sqrt{3}+2+\sqrt{3}=4
\\9)\dfrac{\sqrt{3-\sqrt 5}(3+\sqrt 5)}{\sqrt{10}+\sqrt 2}=\dfrac{\sqrt{3-\sqrt 5}.\sqrt{3+\sqrt 5}.\sqrt{6+2\sqrt 5}}{2(\sqrt 5+1)}=\dfrac{2\sqrt{(\sqrt 5+1)^2}}{2(\sqrt 5+1)}=1
\\12)\sqrt{4-\sqrt{9+4\sqrt{2}}}=\sqrt{4-\sqrt{(2\sqrt{2}+1)^2}}=\sqrt{3-2\sqrt{2}}=\sqrt{(\sqrt{2}-1)^2}=\sqrt{2}-1
\\15)\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}=\dfrac{\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}}{\sqrt{2}}
\\=\dfrac{\sqrt{(\sqrt{5}-1)^2}+\sqrt{(\sqrt{5}+1)^2}}{\sqrt{2}}=\dfrac{\sqrt{5}-1+\sqrt{5}+1}{\sqrt{2}}=\dfrac{2\sqrt 5}{\sqrt 2}=\sqrt{10}
\\18)\dfrac{2}{3+\sqrt 5}+\dfrac{2}{3-\sqrt 5}=\dfrac{2(3-\sqrt 5)+2(3+\sqrt 5)}{9-5}=\dfrac{6-2\sqrt 5+6+2\sqrt 5}{4}=3
\\21)(\sqrt{2}+1)^3-(\sqrt{2}-1)^3=2\sqrt{2}+6+3\sqrt{2}+1-2\sqrt{2}+6-3\sqrt{2}+1=14
\\24)\dfrac{\sqrt{18}}{\sqrt 2}-\dfrac{\sqrt{12}}{\sqrt 3}=\dfrac{\sqrt{36}}{2}-\dfrac{\sqrt{36}}{3}=\dfrac{6}{2}-\dfrac{6}{3}=3-2=1
\\27)\sqrt{3-2\sqrt{2}}=\sqrt{2-2\sqrt{2}+1}=\sqrt{(\sqrt{2}-1)^2}=\sqrt{2}-1$
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
