Đáp án+Giải thích các bước giải:
$\displaystyle \begin{array}{{>{\displaystyle}l}} Bài\ 2:\ 3cos^{2} \alpha -4sin^{2} \alpha \\ \ \ \ \ \ \ \ \ =3\left( 1-sin^{2} \alpha \right) -4sin^{2} \alpha \\ \ \ \ \ \ \ \ \ =3-7sin^{2} \alpha =3-7\left(\frac{1}{5}\right)^{2} =\frac{68}{25}\\ \\ Bài\ 3:\frac{2cos^{2} \alpha -1}{sin\alpha +cos\alpha } =\frac{2cos^{2} \alpha -sin^{2} \alpha -cos^{2} \alpha }{sin\alpha +cos\alpha }\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{cos^{2} \alpha -sin^{2} \alpha }{sin\alpha +cos\alpha }\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{ \begin{array}{{>{\displaystyle}l}} ( sin\alpha -cos\alpha )( sin\alpha +cos\alpha ) \end{array}}{sin\alpha +cos\alpha }\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =sin\alpha -cos\alpha \end{array}$
