[TEX](log_2(x))^2+\sqrt{log_2(x)+1}=1[/TEX]
giai thich >de bai la :loga co so 2 cua x tat ca binh +can bac 2 cua loga co so 2 cua x +1 =1

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[TEX]\sqrt {\log _2 x + 1} + \log _2^2 x - 1 = 0[/TEX]
[TEX]\Rightarrow \sqrt {\log _2 x + 1} \left[ {1 + \sqrt {\log _2 x + 1} \left( {\log _2 x - 1} \right)} \right] = 0[/TEX]
+++++++++ [TEX]\log _2 x + 1 = 0 \Rightarrow x = \frac{1}{2}[/TEX]
+++++++++[TEX]1 + \sqrt {\log _2 x + 1} \left( {\log _2 x - 1} \right){\rm{ = 0 (1)}}[/TEX]
[TEX]\sqrt {\log _2 x + 1} = t \ge 0 \Rightarrow {\rm{ (1)}} \Leftrightarrow {\rm{t}}\left( {t^2 - 1} \right) + 1 = 0[/TEX]
[TEX]\Rightarrow t = 1{\rm{ }};{\rm{ t}} = {\rm{ - }}\frac{{1 - \sqrt 5 }}{2}[/TEX]
[TEX] \Rightarrow x = 1;{\rm{ x}} = 2^{\frac{{1 - \sqrt 5 }}{2}} [/TEX]