\(\text{a) }-9x^2+12x-17\\ =-9x^2+12x-4-13\\ =-\left(9x^2-12x+4\right)-13\\ =-\left[\left(3x\right)^2-2\cdot3x\cdot2+2^2\right]-13\\ =-\left(3x-2\right)^2-13\\ \text{Ta có : }\left(3x-2\right)^2\ge0\forall x\\ \Rightarrow-\left(3x-2\right)^2\le0\forall x\\ \Rightarrow-\left(3x-2\right)^2-13\le13\forall x\\ \Rightarrow-\left(3x-2\right)^2-13< 0\forall x\\ \text{Vậy biểu thức luôn nhận giá trị âm }\forall x\)
\(\text{b) }-11-\left(x-1\right)\left(x+2\right)\\ =-11-x\left(x+2\right)+\left(x+2\right)\\ =-11-x^2-2x+x+2\\ =-\left(11-2\right)-x^2-\left(2x-x\right)\\ =-9-x^2-x\\ =-x^2-x-9\\ =-x^2-x-\dfrac{1}{4}-\dfrac{35}{4}\\ =-\left(x^2+x+\dfrac{1}{4}\right)-\dfrac{35}{4}\\ =-\left[x^2+2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]-\dfrac{35}{4}\\ -\left(x+\dfrac{1}{2}\right)^2-\dfrac{35}{4}\\ \text{Ta có : }\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\\ \Rightarrow-\left(x+\dfrac{1}{2}\right)^2\le0\forall x\\ \Rightarrow-\left(x+\dfrac{1}{2}\right)^2-\dfrac{35}{4}\le-\dfrac{35}{4}\forall x\\ \Rightarrow-\left(x+\dfrac{1}{2}\right)^2-\dfrac{35}{4}< 0\forall x\\ \text{Vậy biểu thức luôn nhận giá trị âm }\forall x\)
có vẻ đề sai bạn ơi
mik chỉ nói zậy chứ ko chắk lắm :x :x
