\(\text{a) }-9x^2+12x-17\\ =-9x^2+12x-4-13\\ =-\left(9x^2-12x+4\right)-13\\ =-\left[\left(3x\right)^2-2\cdot3x\cdot2+2^2\right]-13\\ =-\left(3x-2\right)^2-13\\ \text{Ta có: }\left(3x-2\right)^2\ge0\forall x\\ \Rightarrow-\left(3x-2\right)^2\le0\forall x\\ \Rightarrow-\left(3x-2\right)^2-13\le-13\forall x\\ \Rightarrow\left(3x-2\right)^2-13< 0\forall x\\ \text{Vậy biểu thức luôn nhận giá trị âm }\forall x\\ \\ \)

\(\text{b) }-11-\left(x-1\right)\left(x+2\right)\\ =-11-x\left(x+2\right)-\left(x+2\right)\\ =-11-x^2-2x-x-2\\ =-x^2-\left(2x+x\right)-\left(11+2\right)\\ =-x^2-3x-13\\ =-x^2-3x-\dfrac{9}{4}-\dfrac{43}{4}\\ =-\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{43}{4}\\ =-\left[x^2+2\cdot x\cdot\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2\right]-\dfrac{43}{4}\\ =-\left(x+\dfrac{3}{2}\right)^2-\dfrac{43}{4}\\ \text{Ta có: }\left(x+\dfrac{3}{2}\right)^2\ge0\forall x\\ \Rightarrow-\left(x+\dfrac{3}{2}\right)^2\le0\forall x\\ \Rightarrow-\left(x+\dfrac{3}{2}\right)^2-\dfrac{43}{4}\le-\dfrac{43}{4}\forall x\\ \Rightarrow-\left(x+\dfrac{3}{2}\right)^2-\dfrac{43}{4}< 0\forall x\\ \text{Vậy biểu thức giá trị âm }\forall x\)