a) A= [tex]\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}[/tex]
b) B= [tex]2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}[/tex]
b) C= [tex]\sqrt{m+2\sqrt{m-1}}+\sqrt{m-2\sqrt{m-1}}[/tex]
a) [tex]A=\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}[/tex]
[tex]A=\sqrt{13+30\sqrt{2+\sqrt{8+2.2\sqrt{2}.1+1}}}[/tex]
[tex]A=\sqrt{13+30\sqrt{2+\sqrt{(2\sqrt{2}+1)^2}}}[/tex]
[tex]A=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}[/tex]
[tex]A=\sqrt{13+30\sqrt{(\sqrt{2}+1)^2}}[/tex]
[tex]A=\sqrt{13+30(\sqrt{2}+1)}[/tex]
[tex]A=\sqrt{43+30\sqrt{2}}[/tex]
[tex]A=\sqrt{25+2.5.3\sqrt{2}+18}[/tex]
[tex]A=\sqrt{(5+3\sqrt{2})^2}[/tex]
[tex]A=5+3\sqrt{2}[/tex]
b) [tex]B=2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}[/tex]
[tex]B=2\sqrt{3+\sqrt{5-\sqrt{13+4.\sqrt{3}}}}[/tex]
[tex]B=2\sqrt{3+\sqrt{5-\sqrt{(2\sqrt{3}+1)^2}}}[/tex]
[tex]B=2\sqrt{3+\sqrt{4-2\sqrt{3}}}[/tex]
[tex]B=2\sqrt{3+\sqrt{(\sqrt{3}-1)^2}}[/tex]
[tex]B=2\sqrt{2+\sqrt{3}}=[/tex] $\sqrt{2}.\sqrt{4+2\sqrt{3}}$
[tex]B=\sqrt{2}.(\sqrt{3}+1)[/tex]
c) $DK:..$
[tex]C=\sqrt{m+2\sqrt{m-1}}+\sqrt{m-2\sqrt{m-1}}[/tex]
[tex]C=\sqrt{(\sqrt{m-1}+1)^2}+\sqrt{(\sqrt{m-1}-1)^2}[/tex]
[tex]C=\begin{vmatrix} \sqrt{m-1}+1 \end{vmatrix}+\begin{vmatrix} \sqrt{m-1}-1 \end{vmatrix}[/tex]
[tex]C=\sqrt{m-1}+1+\begin{vmatrix} \sqrt{m-1}-1 \end{vmatrix}[/tex]
Đến đây xét: [tex]\sqrt{m-1}>1;\sqrt{m-1}<1[/tex] kết hợp với $DK$ để bỏ trị tuyệt đối rồi tính!
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
