B1:
a, \(\dfrac{3x+7}{2+x-x^2}\ge5\)
<=> \(\dfrac{3x+7-5\left(2+x-x^2\right)}{2+x-x^2}\ge0\)
<=> \(\dfrac{5x^2+8x-3}{2+x-x^2}\ge0\)
\(5x^2+2x-3=0< =>\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=-1\end{matrix}\right.\)
\(-x^2+x+2=0< =>\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
- Sau đó lập bảng xét dấu và kết luận
B2:
Vì \(\dfrac{\Pi}{2}< x< \Pi\) => \(\cos\alpha< 0\), \(\sin\alpha>0\)
\(\cos2\alpha=1-2\sin^2\alpha=1-2.\left(\dfrac{4}{5}\right)^2=\dfrac{-7}{25}\)\(\cos\alpha=-\sqrt{1-\sin^2\alpha}=-\sqrt{1-\left(\dfrac{4}{5}\right)^2}=\dfrac{-3}{5}\)\(\sin2\alpha=2\sin\alpha.\cos\alpha=2.\dfrac{4}{5}.\left(\dfrac{-3}{5}\right)=\dfrac{-24}{25}\)
