Trả lời:
Đáp án đúng: A
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.
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18/09/2025
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