Đáp án:
B=\(
\frac{{\sqrt 2 - 1}}{{\frac{8}{3}\sqrt 2 + 1}}
\)
Giải thích các bước giải:
\(
\begin{array}{l}
B = \frac{{\sin \alpha - \cos \alpha }}{{\sin ^3 \alpha + 3\cos ^3 \alpha + 2\sin \alpha }} = \frac{{\sqrt 2 \cos \alpha - \cos \alpha }}{{(\sqrt 2 \cos \alpha )^3 + 3\cos ^3 \alpha + 2\sqrt 2 \cos \alpha }} \\
= \frac{{\left( {\sqrt 2 - 1} \right)\cos \alpha }}{{2\sqrt 2 \cos ^3 \alpha + 3\cos ^3 \alpha + 2\sqrt 2 \cos \alpha }} = \frac{{(\sqrt 2 - 1)\cos \alpha }}{{(2\sqrt 2 \cos ^2 \alpha + 3\cos ^2 \alpha + 2\sqrt 2 )\cos \alpha }} \\
= \frac{{\sqrt 2 - 1}}{{2\sqrt 2 .(\frac{{\sqrt 3 }}{3})^2 + 3.\left( {\frac{{\sqrt 3 }}{3}} \right)^2 + 2\sqrt 2 }} = \frac{{\sqrt 2 - 1}}{{\frac{8}{3}\sqrt 2 + 1}} \\
\end{array}
\)
