Chứng minh phát biểu sau đúng bầng tiếng anh

To prove the statement, let's first express the unit vectors \(\mathbf{a}\) and \(\mathbf{b}\) in terms of their angles with the x-axis.

1. **Definition of Unit Vectors**:
- Since \(\mathbf{a}\) is inclined at \(30^\circ\):
\[
\mathbf{a} = \begin{pmatrix} \cos(30^\circ) \\ \sin(30^\circ) \end{pmatrix} = \begin{pmatrix} \frac{\sqrt{3}}{2} \\ \frac{1}{2} \end{pmatrix}
\]

- Since \(\mathbf{b}\) is inclined at \(120^\circ\):
\[
\mathbf{b} = \begin{pmatrix} \cos(120^\circ) \\ \sin(120^\circ) \end{pmatrix} = \begin{pmatrix} -\frac{1}{2} \\ \frac{\sqrt{3}}{2} \end{pmatrix}
\]

2. **Summing the Vectors**:
Now, we compute \(\mathbf{a} + \mathbf{b}\):
\[
\mathbf{a} + \mathbf{b} = \begin{pmatrix} \frac{\sqrt{3}}{2} \\ \frac{1}{2} \end{pmatrix} + \begin{pmatrix} -\frac{1}{2} \\ \frac{\sqrt{3}}{2} \end{pmatrix} = \begin{pmatrix} \frac{\sqrt{3}}{2} - \frac{1}{2} \\ \frac{1}{2} + \frac{\sqrt{3}}{2} \end{pmatrix}
\]
\[
= \begin{pmatrix} \frac{\sqrt{3} - 1}{2} \\ \frac{1 + \sqrt{3}}{2} \end{pmatrix}
\]

3. **Magnitude of the Resultant Vector**:
The magnitude \(|\mathbf{a} + \mathbf{b}|\) is given by the formula:
\[
|\mathbf{a} + \mathbf{b}| = \sqrt{\left( \frac{\sqrt{3} - 1}{2} \right)^2 + \left( \frac{1 + \sqrt{3}}{2} \right)^2}
\]

Computing each term:
\[
= \sqrt{\frac{(\sqrt{3} - 1)^2 + (1 + \sqrt{3})^2}{4}} = \frac{1}{2} \sqrt{(\sqrt{3} - 1)^2 + (1 + \sqrt{3})^2}
\]

4. **Expanding the Squares**:
\[
(\sqrt{3} - 1)^2 = 3 - 2\sqrt{3} + 1 = 4 - 2\sqrt{3}
\]
\[
(1 + \sqrt{3})^2 = 1 + 2\sqrt{3} + 3 = 4 + 2\sqrt{3}
\]

Adding these:
\[
(4 - 2\sqrt{3}) + (4 + 2\sqrt{3}) = 8
\]

5. **Final Calculation**:
Now substituting back to find the magnitude:
\[
|\mathbf{a} + \mathbf{b}| = \frac{1}{2} \sqrt{8} = \frac{1}{2} \cdot 2\sqrt{2} = \sqrt{2}
\]

Thus, the value of \(|\mathbf{a} + \mathbf{b}|\) is \(\sqrt{2}\).