Cho \(tanx = m\). Giá trị của \(\frac{{{\mathop{\rm sinx}\nolimits}  - c{\rm{osx}}}}{{2{{\sin }^3}x - c{\rm{osx}}}}\) bằng

* Hướng dẫn giải

\(\begin{array}{l}
\frac{{\sin x - c{\rm{osx}}}}{{2{\mathop{\rm s}\nolimits} {\rm{i}}{{\rm{n}}^3}x - c{\rm{osx}}}} = \frac{{(\sin x - c{\rm{osx}})({{\sin }^2}x + c{\rm{o}}{{\rm{s}}^2}{\rm{x)}}}}{{2{{\sin }^3}x - c{\rm{osx(}}{{\sin }^2}x + c{\rm{o}}{{\rm{s}}^2}{\rm{x}})}}\\
 = \frac{{\left( {\frac{{{\mathop{\rm s}\nolimits} {\rm{inx}}}}{{{\rm{cosx}}}} - 1} \right)\left( {\frac{{{\mathop{\rm s}\nolimits} {\rm{i}}{{\rm{n}}^2}{\rm{x}}}}{{{\rm{co}}{{\rm{s}}^2}{\rm{x}}}} + 1} \right)}}{{\frac{{2{{\sin }^3}x}}{{{{\cos }^3}x}} - \left( {\frac{{{\mathop{\rm s}\nolimits} {\rm{i}}{{\rm{n}}^2}{\rm{x}}}}{{{\rm{co}}{{\rm{s}}^2}{\rm{x}}}} + 1} \right)}} = \frac{{{\rm{(tanx - 1)(ta}}{{\rm{n}}^2}{\rm{x + 1)}}}}{{2{{\tan }^3}x - {\rm{(ta}}{{\rm{n}}^2}{\rm{x + 1)}}}}\\
 = \frac{{(m - 1)({m^2} + 1)}}{{2{m^3} - {m^2} - 1}} = \frac{{{m^2} + 1}}{{2{m^2} + m + 1}}.
\end{array}\)