Bài 1
[tex]a,A=\sqrt{13+\sqrt{30\sqrt{2+\sqrt{9+4\sqrt{2}}}}}=\sqrt{13+\sqrt{30\sqrt{2+(1+\sqrt{8})}}}=\sqrt{13+\sqrt{30(1+\sqrt{2})}}[/tex] (không biết bạn có chép sai đề câu này không nhỉ?)
[tex]b,B=\sqrt{|40\sqrt{2}-57|}-\sqrt{|40\sqrt{2}+57|}=\sqrt{-40\sqrt{2}+57}-\sqrt{40\sqrt{2}+57}[/tex]
[tex]\Rightarrow B^2=-40\sqrt{2}+57+40\sqrt{2}+57-2\sqrt{-40\sqrt{2}+57}.\sqrt{40\sqrt{2}+57}=114-2\sqrt{49}=100[/tex]
[tex]\Rightarrow B=-10(do.B<0)[/tex]
[tex]c,C=(2+\sqrt{3}-\sqrt{2})(2-\sqrt{3}-\sqrt{2})(3+\sqrt{2})\sqrt{3-2\sqrt{2}}=[(2-\sqrt{2})^2-3](3+\sqrt{2})\sqrt{(\sqrt{2}-1)^2}=(3-4\sqrt{2})(3+\sqrt{2})(\sqrt{2}-1)=10\sqrt{2}-19[/tex]
[tex]d,D=(4\sqrt{2}+30)(\sqrt{5}-\sqrt{3})\sqrt{4-\sqrt{15}}=\sqrt{2}(2\sqrt{2}+15)(\sqrt{5}-\sqrt{3})\sqrt{8-2\sqrt{15}}[/tex][tex]=\sqrt{2}(2\sqrt{2}+15)(\sqrt{5}-\sqrt{3})\sqrt{8-2\sqrt{15}}=(4+15\sqrt{2})(\sqrt{5}-\sqrt{3})^2=(4+15\sqrt{2})(8-2\sqrt{15})=32+120\sqrt{2}-8\sqrt{15}-30\sqrt{30}[/tex]
Bài 2
[tex]\sqrt{21-6\sqrt{6}}+\sqrt{9+2\sqrt{48}}-2\sqrt{6+3\sqrt{3}}\approx 0.59929[/tex] (xem lại đề)
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
