$1)f(x)=\sqrt{\dfrac{x^2}{2}-2x+4}\\ f'(x)=\dfrac{\left(\dfrac{x^2}{2}-2x+4\right)'}{2\sqrt{\dfrac{x^2}{2}-2x+4}}\\ =\dfrac{x-2}{2\sqrt{\dfrac{x^2}{2}-2x+4}}\\ 2)y=\left\{\begin{array}{cc} \dfrac{x^2-16}{x-4}&x\ne 4\\ 2a&x=4\end{array} \right.\\ \displaystyle\lim_{x \to 4} f(x)\\ =\displaystyle\lim_{x \to 4} \dfrac{x^2-16}{x-4}\\ =\displaystyle\lim_{x \to 4} \dfrac{(x-4)(x+4)}{x-4}\\ =\displaystyle\lim_{x \to 4} x+4\\ =4+4\\ =8\\ \text{f(x) liên tục tại} x=4 \Leftrightarrow f(4)=\displaystyle\lim_{x \to 4} f(x)\\ \Leftrightarrow 2a=8\\ \Leftrightarrow a=4\\ 3)y=2x^5-\dfrac{4}{x}+5\\ y'=10x^4+\dfrac{4}{x^2}\\ y'(-1)=10.(-1)^4+\dfrac{4}{(-1)^2}=14\\ \Rightarrow B$