[imath]\sqrt{5+4 \sqrt{7-2 \sqrt{6- \sqrt{32}}}} \\
=\sqrt{5+4 \sqrt{7-2 \sqrt{6- 4 \sqrt{2}}}} \\
=\sqrt{5+4 \sqrt{7-2 \sqrt{4- 2 . \sqrt{2}.2+2}}} \\
=\sqrt{5+4 \sqrt{7-2 \sqrt{(2- \sqrt{2})^2}}}[/imath]
[imath]=\sqrt{5+4 \sqrt{7-2 (2- \sqrt{2})}}[/imath] (vì [imath]2> \sqrt{2}[/imath] nên [imath]\sqrt{(2- \sqrt{2})^2} = |2- \sqrt{2}| = 2 - \sqrt{2}[/imath])
[imath]=\sqrt{5+4 \sqrt{3+2 \sqrt{2}}} \\
=\sqrt{5+4 \sqrt{2+2 . \sqrt{2}.1+1}} \\
=\sqrt{5+4 \sqrt{(\sqrt{2}+1)^2}} \\
=\sqrt{5+4(\sqrt{2}+1)} \\
=\sqrt{9+4 \sqrt{2}} \\
=\sqrt{1+2. 2 \sqrt{2}.1+8} \\
=\sqrt{(1+2 \sqrt{2})^2} \\
=1+2 \sqrt{2} = \sqrt{1} + \sqrt{8} [/imath]
Vì [imath]a<b[/imath] nên [imath]a=1, \ b=8[/imath]
Vậy [imath]T=1^2+8=9[/imath]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
