(1,0 điểm) Cho \(\alpha \) là góc nhọn và \({\rm{sin}}\frac{\alpha }{2} = \sqrt {\frac{{x - 1}}{{2x}}} \). Tìm \(x\) để \({\rm{tan}}\alpha = \frac{1}{2}x\).

Ta có: \(0 < \alpha < 90^\circ \Leftrightarrow 0 < \frac{\alpha }{2} < 45^\circ \Rightarrow 0 < {\rm{sin}}\frac{\alpha }{2} < \frac{{\sqrt 2 }}{2} \Leftrightarrow 0 < \sqrt {\frac{{x - 1}}{{2x}}} < \frac{{\sqrt 2 }}{2} \Leftrightarrow x > 0\)

Lại có \({\rm{si}}{{\rm{n}}^2}\frac{\alpha }{2} + {\rm{co}}{{\rm{s}}^2}\frac{\alpha }{2} = 1 \Rightarrow {\rm{cos}}\frac{\alpha }{2} = \sqrt {1 - {\rm{si}}{{\rm{n}}^2}\frac{\alpha }{2}} \), vì \(0 < \frac{\alpha }{2} < 45^\circ \)

\( \Leftrightarrow {\rm{cos}}\frac{\alpha }{2} = \sqrt {\frac{{x + 1}}{{2x}}} \Rightarrow {\rm{tan}}\frac{\alpha }{2} = \sqrt {\frac{{x - 1}}{{x + 1}}} \)

Khi đó \({\rm{tan}}\alpha = \frac{{2{\rm{tan}}\frac{\alpha }{2}}}{{1 - {\rm{ta}}{{\rm{n}}^2}\frac{\alpha }{2}}} = \frac{{2\sqrt {\frac{x}{{x + 1}}} }}{{1 - \frac{{x - 1}}{{x + 1}}}} = \sqrt {{x^2} - 1} \).

Ta có: \({\rm{tan}}\alpha = \frac{1}{2}x\) \( \Leftrightarrow \sqrt {{x^2} - 1} = \frac{1}{2}x\)

\( \Leftrightarrow {x^2} - 1 = \frac{1}{4}{x^2}\) (do \(x > 0\))

\( \Leftrightarrow  - \frac{3}{4}{x^2} = - 1\)

\( \Leftrightarrow {x^2} = \frac{4}{3}\)\( \Leftrightarrow x = \frac{2}{{\sqrt 3 }} = \frac{{2\sqrt 3 }}{3}\) (do \(x > 0\))

Vậy giá trị \(x\) cần tìm là \(x = \frac{{2\sqrt 3 }}{3}\).