Đáp án:

•  M \(\in\) mp \(Oxy \Leftrightarrow M(x,y,0)\)

•  MA = MB = MC

\(\Leftrightarrow 
\begin{cases}
MA = MB \\
MA = MC
\end{cases}
\Leftrightarrow 
\begin{cases}
(x-1)^2 + (y+1)^2 + (-5)^2 = (x-3)^2 + (y-4)^2 + (-4)^2 \\
(x-1)^2 + (y+1)^2 + (-5)^2 = (x-4)^2 + (y-6)^2 + (-1)^2
\end{cases}\)

\(\Leftrightarrow 
\begin{cases}
-2x + 2y + 27 = -6x - 8y + 41 \\
-2x + 2y + 27 = -8x - 12y + 53
\end{cases}\)

\(\Leftrightarrow 
\begin{cases}
4x + 10y = 14 \\
6x + 14y = 26
\end{cases}
\Leftrightarrow 
\begin{cases}
x = 16 \\
y = -5
\end{cases}\)

\(\Rightarrow\) Vậy chọn \(\boxed{\text{A}} \)

page 25

Đáp án:

•  \(|\overrightarrow{p}|^2 = \overrightarrow{p}^2 = (3\overrightarrow{a} - 2\overrightarrow{b})^2 \)
= \(9\overrightarrow{a}^2 + 4\overrightarrow{b}^2 - 12\overrightarrow{a}. \overrightarrow{b}\)

= \( 36 + 36 - 12 . |\overrightarrow{a}| . |\overrightarrow{b}| \cos(\overrightarrow{a}, \overrightarrow{b})\)

= \(72 - 12 . 2 . 3 .\frac{1}{2} = 72 - 36 = 36\)

\(\Rightarrow |\overrightarrow{p}| = 6\)

\(\Rightarrow\) Vậy chọn \(\boxed{\text{A}} \)

page 26

Đáp án:

•  \(|\overrightarrow{p}|^2 = \overrightarrow{p}^2 = 4|\overrightarrow{a}|^2 + 9|\overrightarrow{b}|^2 + |\overrightarrow{c}|^2 - 12\overrightarrow{ab} - 6\overrightarrow{bc} + 4\overrightarrow{ac}\)

                      \(= 4 + 36 + 9 = 49 \Rightarrow |\overrightarrow{p}| = 7\)

\(\Rightarrow\) Vậy chọn \(\boxed{\text{B}} \)

page 27

​​​​​​​Đáp án:

•  \( \overrightarrow{p} . \overrightarrow{q} \) \( = (\overrightarrow{a} - \overrightarrow{b}) . (\overrightarrow{a} + 2\overrightarrow{b}) = \overrightarrow{a}^2 - 2\overrightarrow{b}^2 + \overrightarrow{a}.\overrightarrow{b}\)

                 \(= 4-18 = -14\)

•  \(\overrightarrow{p}^2 = (\overrightarrow{a} - \overrightarrow{b})^2 = \overrightarrow{a}^2 - 2\overrightarrow{a} \overrightarrow{b} + \overrightarrow{b}^2 = 4 + 9 = 13 \quad \Rightarrow |\overrightarrow{p}| = \sqrt{13}
\)

•  \(\overrightarrow{q}^2 = (\overrightarrow{a} + 2\overrightarrow{b})^2 = \overrightarrow{a}^2 + 4\overrightarrow{b} + 4\overrightarrow{a} .\overrightarrow{b}^2 = 4 + 36 = 40 \quad \)

\(\Rightarrow |\overrightarrow{q}| = 2\sqrt{10}
\)

•  \(\overrightarrow{p}. \overrightarrow{q} = |\overrightarrow{p}| .|\overrightarrow{q}| \cos(\overrightarrow{p}, \overrightarrow{q})\)

\(\Rightarrow\cos(\overrightarrow{p}, \overrightarrow{q}) = \frac{\overrightarrow{p}. \overrightarrow{q}}{|\overrightarrow{p}| |\overrightarrow{q}|} = \frac{-14}{\sqrt{13}  .2\sqrt{10}} = -\frac{7\sqrt{130}}{130}
\)

\(\Rightarrow\) Vậy chọn \(\boxed{\text{A}} \)

page 28

​​​​​​​Đáp án:

•  \(\vec{a} - \vec{b} = (x - 2, 0, 1)\), \(\vec{a} - \vec{c} = (x - 3, -1, 0)\)

•  \(P = |\vec{a} - \vec{b}| + |\vec{a} - \vec{c}| = \sqrt{(x - 2)^2 + 1} + \sqrt{(x - 3)^2 + 1}\)

\(= \sqrt{x^2 - 4x + 5} + \sqrt{x^2 - 6x + 10} = f(x)\)

Thử:  
•  \(x = 2 \quad \Rightarrow \quad f(x) = \sqrt{5} + \sqrt{2} \approx 2.4142\)  
•  \(x = 3 \quad \Rightarrow \quad f(x) = 1 + 2\sqrt{2} \approx 2.4142\)  
•  \(x = \frac{5}{2} \quad \Rightarrow \quad f(x) = \frac{5 + \sqrt{5}}{2} \approx 2.236\)  
•  \(x = \frac{7}{3} \quad \Rightarrow \quad f(x) = \frac{\sqrt{13} + \sqrt{10}}{3} \approx 2.2553\)

\(\Rightarrow\) Vậy chọn \(\boxed{\text{D}} \)

​​​​​​​Đáp án:

Bấm: Mode → 7

\( P = \sqrt{(x-2)^2 + 1} + \sqrt{(x-3)^2 + 1} \)  

•  \(M(x, 0) ∈ Ox  \)

•  \(A(1, 1), B(3, 1 )\)

•  \( P = MA + MB \) nhỏ nhất \(\Leftrightarrow\) M ≡ I  

\(\Leftrightarrow\) \( x = \frac{5}{2} \)

​​​​​​​Đáp án:

\(\Rightarrow\) Vậy chọn \(\boxed{\text{A}} \)

Hoặc: \( \left( \frac{2}{5} x \right)^3 + \frac{x}{5} \) bằng  

A. 1              B. 2              C. 3              D. \( \frac{2}{5} \)

\(\Rightarrow\) Vậy chọn \(\boxed{\text{C}} \)

page 29