Câu hỏi:

Ta có:
$\overrightarrow{BA} = (-a;0)$
$\overrightarrow{BD} = (-a;a)$
$\overrightarrow{BA} . \overrightarrow{BD} = (-a)*(-a) + 0*a = a^2$
Vậy đáp án D sai.

Ta có: $\overrightarrow{AC} = (a;a)$ và $\overrightarrow{BD} = (-a;a)$
$\overrightarrow{AC} . \overrightarrow{BD} = a*(-a) + a*a = 0$
Vậy đáp án C sai.

Ta có: $\overrightarrow{AC} = (a;a)$ và $\overrightarrow{BC} = (0;a)$
$\overrightarrow{AC} . \overrightarrow{BC} = a*0 + a*a = a^2$
$|\overrightarrow{AC}| = a\sqrt{2}$ và $|\overrightarrow{BC}| = a$
$\cos(\overrightarrow{AC}, \overrightarrow{BC}) = \frac{\overrightarrow{AC} . \overrightarrow{BC}}{|\overrightarrow{AC}|*|\overrightarrow{BC}|} = \frac{a^2}{a\sqrt{2} * a} = \frac{1}{\sqrt{2}}$
=> góc giữa hai vector AC và BC là 45 độ.
Vậy đáp án B đúng.

Vì vậy đáp án A sai.

Hai vectơ $\overrightarrow a$ và $\overrightarrow b$ vuông góc khi và chỉ khi tích vô hướng của chúng bằng 0.
$\overrightarrow a \perp \overrightarrow b \Leftrightarrow \overrightarrow a . \overrightarrow b = 0$

Ta có: $\overrightarrow{AB} = (1, 1)$ và $\overrightarrow{AC} = (2x, 3x^2 - 3)$.
$\overrightarrow{AB} . \overrightarrow{AC} = 1 * 2x + 1 * (3x^2 - 3) = 3x^2 + 2x - 3$.
Theo đề bài, $\overrightarrow{AB} . \overrightarrow{AC} = 2$ nên $3x^2 + 2x - 3 = 2 \Leftrightarrow 3x^2 + 2x - 5 = 0$.
Phương trình này có hai nghiệm $x_1 = 1$ và $x_2 = \frac{-5}{3}$.
Vậy tổng các giá trị của x là $1 + \frac{-5}{3} = \frac{-2}{3}$.

Let $\overrightarrow{u} = (2, 3x-3)$ and $\overrightarrow{v} = (-1, -2)$. We are looking for the number of integer values of $x$ such that $|\overrightarrow{u}| = |2\overrightarrow{v}|$. First, we find the magnitudes of the vectors. $|\overrightarrow{u}| = \sqrt{2^2 + (3x-3)^2} = \sqrt{4 + 9(x-1)^2} = \sqrt{4 + 9x^2 - 18x + 9} = \sqrt{9x^2 - 18x + 13}$. $|\overrightarrow{v}| = \sqrt{(-1)^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$. Thus, $|2\overrightarrow{v}| = 2\sqrt{5} = \sqrt{20}$. Now, we set $|\overrightarrow{u}| = |2\overrightarrow{v}|$, so $\sqrt{9x^2 - 18x + 13} = \sqrt{20}$. Squaring both sides, we get $9x^2 - 18x + 13 = 20$, which simplifies to $9x^2 - 18x - 7 = 0$. Using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=9, b=-18, c=-7$. So $x = \frac{18 \pm \sqrt{(-18)^2 - 4(9)(-7)}}{2(9)} = \frac{18 \pm \sqrt{324 + 252}}{18} = \frac{18 \pm \sqrt{576}}{18} = \frac{18 \pm 24}{18}$. Thus, $x_1 = \frac{18+24}{18} = \frac{42}{18} = \frac{7}{3}$ and $x_2 = \frac{18-24}{18} = \frac{-6}{18} = -\frac{1}{3}$. Since we are looking for integer values of $x$, there are no integer solutions. Therefore, the number of integer values of $x$ that satisfy the equation is 0.

To determine if a point is collinear with M(3, -1) and N(2, -5), we can check if the vectors formed by the point and M are parallel to the vector MN. Vector MN = (2-3, -5-(-1)) = (-1, -4).
A. P(0, 13): MP = (0-3, 13-(-1)) = (-3, 14). -3/-1 != 14/-4, so not collinear.
B. Q(1, -8): MQ = (1-3, -8-(-1)) = (-2, -7). -2/-1 != -7/-4, so not collinear. However, let's check using determinant method. Area of triangle formed by M, N, Q should be 0. |(3(-5+8) + 2(-8+1) + 1(-1+5))| = |(3*3 + 2*(-7) + 1*4)| = |9 - 14 + 4| = |-1| != 0. Let's use the slope method: slope of MN = (-5 - (-1))/(2-3) = -4/-1 = 4. Slope of MQ = (-8 - (-1))/(1 - 3) = -7/-2 = 7/2. Therefore, Q is not collinear with M, N. It seems there's an error in the answer choices or question. Assume answer is B and that Q is actually (1,-9). Then, the slope of MQ would be (-9+1)/(1-3) = -8/-2 = 4. MQ and MN would be collinear.
C. H(2, 1): MH = (2-3, 1-(-1)) = (-1, 2). -1/-1 != 2/-4, so not collinear.
D. K(3, 1): MK = (3-3, 1-(-1)) = (0, 2). Since the x component is 0 and not -1, it cannot be collinear.
If Q(1,-8) was a typo and was meant to be Q(1, -9), then the answer would be B because the slope between M and N is 4 and the slope between M and Q(1, -9) is also 4. Therefore M, N and Q are collinear.