To solve the equations and systems given in the question, let's go through them one by one.

### Bài 1: Giải phương trình, hệ phương trình

#### a) \(3(x-5)(x+2) = x^2 - 5x\)

1. Expand the left side:
\[
3(x^2 - 3x - 10) = x^2 - 5x
\]
\[
3x^2 - 9x - 30 = x^2 - 5x
\]

2. Rearranging gives:
\[
3x^2 - 9x - 30 - x^2 + 5x = 0
\]
\[
2x^2 - 4x - 30 = 0
\]
\[
x^2 - 2x - 15 = 0
\]

3. Factor or use the quadratic formula:
\[
(x-5)(x+3) = 0 \quad \Rightarrow \quad x = 5 \text{ or } x = -3
\]

---

#### b) Solve the system:

1. \(3x - 12 = 3\) and \(x - 4 = 8 - 2x\)

From \(3x - 12 = 3\):
\[
3x = 15 \quad \Rightarrow \quad x = 5
\]

Substitute \(x = 5\) into the second equation:
\[
5 - 4 = 8 - 2(5)
\]
\[
1 = 8 - 10 \quad \Rightarrow \quad 1 = -2 \text{ (not consistent)}
\]

Therefore, it seems there's an inconsistency.

2. \(2(x+y) - 5y = 3\)
Rearranging gives:
\[
2x - 3y = 3
\]

Combined with previous results, substitute values accordingly.

---

#### c) \(4(x-1) - 2(y+1) = 4\)

Expand:
\[
4x - 4 - 2y - 2 = 4
\]
\[
4x - 2y - 6 = 4
\]
\[
4x - 2y = 10 \quad \Rightarrow \quad 2x - y = 5
\]

---

#### d) \((x-1)^2 + x^2 \leq (x+1)^2 + (x+2)^2\)

Expand both sides and simplify:
\[
(x-1)^2 + x^2 = x^2 - 2x + 1 + x^2 = 2x^2 - 2x + 1
\]
\[
(x+1)^2 + (x+2)^2 = x^2 + 2x + 1 + x^2 + 4x + 4 = 2x^2 + 6x + 5
\]

Set up the inequality:
\[
2x^2 - 2x + 1 \leq 2x^2 + 6x + 5
\]
\[
-8x - 4 \leq 0 \quad \Rightarrow \quad 8x \geq -4
\]
\[
x \geq -\frac{1}{2}
\]

---

#### e) \(\sqrt{3x^2} - \sqrt{12} = 0\)

1. Solve:
\[
\sqrt{3}x = \sqrt{12} \quad \Rightarrow \quad \sqrt{3}x = 2\sqrt{3}
\]
\[
x = 2
\]

### Kết luận:
- a) \(x = 5\) or \(x = -3\)
- b) Inconsistent system
- c) \(2x - y = 5\)
- d) \(x \geq -\frac{1}{2}\)
- e) \(x = 2\)