$3.\\a)A=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\\\Rightarrow A^2=\left (\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}} \right )^2\\=(\sqrt{4+\sqrt{7}})^2-2\sqrt{4+\sqrt{7}}.\sqrt{4-\sqrt{7}}+(\sqrt{4-\sqrt{7}})^2\\=4+\sqrt{7}-2\sqrt{(4+\sqrt{7})(4-\sqrt{7})}+4-\sqrt{7}\\=8-2\sqrt{16-7}=8-2\sqrt{9}=2\\\Rightarrow A=\sqrt{2}(vì \ \sqrt{4+\sqrt{7}}>\sqrt{4-\sqrt{7}})$
$b)B=-\dfrac{\sqrt{\sqrt{7}-\sqrt{3}}-\sqrt{\sqrt{7}+\sqrt{3}}}{\sqrt{7}-2}\\\Rightarrow B=\dfrac{\sqrt{\sqrt{7}+\sqrt{3}}-\sqrt{\sqrt{7}-\sqrt{3}}}{\sqrt{7}-2}\\\Rightarrow B^2=\dfrac{\left (\sqrt{\sqrt{7}+\sqrt{3}}-\sqrt{\sqrt{7}-\sqrt{3}} \right )^2}{(\sqrt{7}-2)^2}\\=\dfrac{\sqrt{7}+\sqrt{3}-2\sqrt{(\sqrt{7}+\sqrt{3})(\sqrt{7}-\sqrt{3})}+\sqrt{7}-\sqrt{3}}{(\sqrt{7}-2)^2}\\=\dfrac{2\sqrt{7}-2\sqrt{7-3}}{(\sqrt{7}-2)^2}=\dfrac{2\sqrt{7}-2.2}{(\sqrt{7}-2)^2}=\dfrac{2(\sqrt{7}-2)}{(\sqrt{7}-2)^2}=\dfrac{2}{\sqrt{7}-2}$
$C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{4+2\sqrt{3}}}}=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{3+2\sqrt{3}+1}}}\\=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{(\sqrt{3}+1)^2}}}=\sqrt{6+2\sqrt{2}.\sqrt{3-(\sqrt{3}+1)}}\\=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{3}-1}}=\sqrt{6+2\sqrt{2}.\sqrt{2-\sqrt{3}}}\\=\sqrt{6+2\sqrt{2(2-\sqrt{3})}}=\sqrt{6+2\sqrt{4-2\sqrt{3}}}=\sqrt{6+2\sqrt{3-2\sqrt{3}+1}}\\=\sqrt{6+2\sqrt{(\sqrt{3}-1)^2}}=\sqrt{6+2(\sqrt{3}-1)}=\sqrt{6+2\sqrt{3}-2}\\=\sqrt{3+2\sqrt{3}+1}=\sqrt{(\sqrt{3}+1)^2}=\sqrt{3}+1$
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
