$1)f(x)=x^3-2x^2+5\\ f'(x)=3x^2-4x\\ f\left(\dfrac{1}{2}\right)=-\dfrac{5}{4}\\ g(x)=\sin\left(x+\dfrac{\pi}{3}\right)-x\\ g'(x)=\cos\left(x+\dfrac{\pi}{3}\right)\\ g''(x)=-\sin\left(x+\dfrac{\pi}{3}\right)\\ g''\left(\dfrac{\pi}{2}\right)=-\dfrac{1}{2}\\ \dfrac{f\left(\dfrac{1}{2}\right)}{g''\left(\dfrac{\pi}{2}\right)}=\dfrac{5}{2}$
$2)$Tiếp điểm $M(x_0;y_0)$
$2y-10x+1=0\\ \Leftrightarrow y=5x-\dfrac{1}{2}$
Tiếp tuyến song song với $y=5x-\dfrac{1}{2}$
$\Rightarrow f'(x_0)=5\\ \Leftrightarrow \dfrac{5}{(x_0+2)^2}=5\\ \Leftrightarrow (x_0+2)^2=1\\ \Leftrightarrow \left\{\begin{array}{l} x_0=-1\Rightarrow y_0=-2\\ x_0=-3\Rightarrow y_0=8\end{array} \right.$
$\Rightarrow M_1(-1;-2);M_1(-3;8)$ là các tiếp điểm
$Pttt:y_1=5(x+1)-2 \Leftrightarrow y=5x+3\\ y_2=5(x+3)+8 \Leftrightarrow y=5x+23$
