Câu hỏi:

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.

Ta có:

• MN là đường trung bình của tam giác ABC nên MN // BC và MN = BC/2
  => \(\overrightarrow {MN} \) cùng hướng với \(\overrightarrow {BC} \) và |\(\overrightarrow {MN} \)| = 1/2 |\(\overrightarrow {BC} \)|
  
• PC = 1/2 BC và \(\overrightarrow {PC} \) ngược hướng với \(\overrightarrow {BC} \)
  => \(\overrightarrow {MN} \) ≠ \(\overrightarrow {PC} \) => A sai
  
• \(\overrightarrow {AA} \) = \(\overrightarrow {0} \), \(\overrightarrow {PP} \) = \(\overrightarrow {0} \) nên \(\overrightarrow {AA} \) và \(\overrightarrow {PP} \) cùng hướng => B đúng
  
• M là trung điểm AB nên \(\overrightarrow {MB} \) = - \(\overrightarrow {AM} \) => \(\overrightarrow {MB} \) = \(\overrightarrow {AM} \) là sai => C đúng
  
• PB = 1/2 BC và \(\overrightarrow {PB} \) ngược hướng với \(\overrightarrow {BC} \)
  => \(\overrightarrow {MN} \) ngược hướng với \(\overrightarrow {PB} \) và |\(\overrightarrow {MN} \)| = |\(\overrightarrow {PB} \)|=1/2 BC => \(\overrightarrow {MN} \) = -\(\overrightarrow {PB} \) => D đúng
  

Trong hình bình hành ABCD, ta có các cạnh đối song song và bằng nhau. Do đó, $\overrightarrow{CD} = \overrightarrow{BA}$. Vậy đáp án đúng là D.

Để ba điểm M, N, và một điểm khác (ví dụ, P) thẳng hàng, vector $\vec{MN}$ và vector $\vec{MP}$ phải cùng phương, tức là tỉ lệ.
Ta có $\vec{MN} = (2-3; -5-(-1)) = (-1; -4)$.
Xét từng đáp án:

• A. P(0; 13): $\vec{MP} = (0-3; 13-(-1)) = (-3; 14)$. Kiểm tra tỉ lệ: $\frac{-3}{-1} = 3 \neq \frac{14}{-4} = -3.5$. Vậy P không thẳng hàng với M, N.
• B. Q(1; -8): $\vec{MQ} = (1-3; -8-(-1)) = (-2; -7)$. Kiểm tra tỉ lệ: $\frac{-2}{-1} = 2 \neq \frac{-7}{-4} = 1.75$. Vậy Q không thẳng hàng với M, N.
• B. Q(1; -8): corrected calculation $\vec{MQ} = (1-3, -8 - (-1)) = (-2, -7)$ This appears to be incorrect in the original calculation. However if we use $\vec{MQ}=(1-3;-8+1)=(-2;-7)$ then the ratio is $\frac{-2}{-1}=2$ and $\frac{-7}{-4}=\frac{7}{4}$ still not equal. However if we instead consider $\vec{NQ}=(1-2;-8+5)=(-1;-3)$ then the ratio between this and $\vec{MN}=(-1;-4)$ is $\frac{-1}{-1}=1$ and $\frac{-3}{-4}=0.75$. Hence not equal. We were supposed to verify if the vectors are scalar multiples not compare the $\vec{MQ}$ from B with $\vec{MN}$. $\frac{-1}{-1}=1 \frac{-3}{-4}=0.75$. Consider calculating gradient between points: gradient between $M(3;-1)$ and $N(2;-5)$ is $\frac{-5+1}{2-3}=4$. The line equation is $y+1=4(x-3)$ so $y=4x-13$.
  • A: $13 \neq 4(0)-13=-13$ not on the line
  • B: $-8 \neq 4(1)-13=-9$ not on the line
  • Incorrect point C and D are impossible to determine and are redundant.
  • However $Q$ is closed $-9 \approx -8$ but is likely incorrect. $4x -13$. $-8 = 4x -13$, $4x=5$ so $x=1.25$ and its close!

Let's instead calculate equation from $M$ to $N$. $M(3;-1)$ and $N(2;-5)$. $rac{y+1}{x-3} = rac{-5+1}{2-3}=4$. So $y=4x-13$. Check B. $Q(1;-8)$, $-8 = 4(1)-13$ so $-8 = -9$. Not correct. However it is $Q(1;-9)$ that is in the line. So $Q$ is more close than A. I think this is wrong. lets check C and D C: H(2;1) gradient with M(3;-1) is $\frac{1+1}{2-3} = -2$ not 4 so is wrong. D: K(3;1), gradient with M is $\frac{1+1}{3-3}$ which is undefined. So is not equal. There is an error. If we test Q(1;-9) Then $\vec{MQ}=(1-3;-9+1)=(-2;-8)$. $\vec{MN}=(-1,-4)$ then ratio is $2$ and $2$. so the answer is $Q(1;-9)$. But $-9$ is not any of the answers.