Giải các phương trình sau

Let's solve each of the equations step by step.

### B)
\[ 3x + 2 - 2x - 3 = \frac{(2x - 3)}{4} - 6x + 9 \]

1. Simplify the left side:
\[ (3x - 2x) + (2 - 3) = x - 1 \]
So, we have:
\[ x - 1 = \frac{(2x - 3)}{4} - 6x + 9 \]

2. Multiply everything by 4 to eliminate the fraction:
\[ 4(x - 1) = 2x - 3 - 24x + 36 \]
\[ 4x - 4 = -22x + 33 \]

3. Combine like terms:
\[ 4x + 22x = 33 + 4 \]
\[ 26x = 37 \]
\[ x = \frac{37}{26} \]

### f)
\[ \frac{x^2}{3} + \frac{x^2}{4} = 0.5x - 2.5 \]

1. Find a common denominator for the left side (which is 12):
\[ \frac{4x^2}{12} + \frac{3x^2}{12} = 0.5x - 2.5 \]
\[ \frac{7x^2}{12} = 0.5x - 2.5 \]

2. Multiplying everything by 12:
\[ 7x^2 = 6x - 30 \]

3. Rearranging gives:
\[ 7x^2 - 6x + 30 = 0 \]

4. Now using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ a = 7, \ b = -6, \ c = 30 \]
\[ x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 7 \cdot 30}}{2 \cdot 7} \]
\[ x = \frac{6 \pm \sqrt{36 - 840}}{14} \]
\[ x = \frac{6 \pm \sqrt{-804}}{14} \]
Since the discriminant is negative, there are no real solutions.

### g)
\[ 2(3x + 1) + 1 - 5 = \frac{2(3x - 1)}{5} \]

1. Simplify the left side:
\[ 6x + 2 + 1 - 5 = 6x - 2 \]
So:
\[ 6x - 2 = \frac{2(3x - 1)}{5} \]

2. Multiply everything by 5 to eliminate the fraction:
\[ 5(6x - 2) = 2(3x - 1) \]
\[ 30x - 10 = 6x - 2 \]

3. Rearranging gives:
\[ 30x - 6x = -2 + 10 \]
\[ 24x = 8 \]
\[ x = \frac{1}{3} \]

### Summary of Solutions:
- B) \( x = \frac{37}{26} \)
- f) No real solutions
- g) \( x = \frac{1}{3} \)