Để tìm số hữu tỉ \( x \) trong các bài toán sau, ta sẽ giải từng phương trình một.

a) \(\left(\frac{2}{5} - x\right) : \frac{1}{3} + \frac{1}{2} = -4\)

Đầu tiên, ta giải phương trình:
\[
\left(\frac{2}{5} - x\right) : \frac{1}{3} + \frac{1}{2} = -4
\]
Chuyển vế:
\[
\left(\frac{2}{5} - x\right) \cdot 3 + \frac{1}{2} = -4
\]
\[
3 \left(\frac{2}{5} - x\right) + \frac{1}{2} = -4
\]
\[
3 \left(\frac{2}{5} - x\right) = -4 - \frac{1}{2}
\]
\[
3 \left(\frac{2}{5} - x\right) = -\frac{9}{2}
\]
\[
\frac{2}{5} - x = -\frac{3}{2}
\]
\[
-x = -\frac{3}{2} - \frac{2}{5}
\]
\[
-x = -\frac{15}{10} - \frac{4}{10}
\]
\[
-x = -\frac{19}{10}
\]
\[
x = \frac{19}{10}
\]

b) \(\left(-3 + \frac{3}{x} - \frac{1}{3}\right) : \left(1 + \frac{2}{5} + \frac{2}{3}\right) = -\frac{5}{4}\)

Đầu tiên, ta giải phương trình:
\[
\left(-3 + \frac{3}{x} - \frac{1}{3}\right) : \left(1 + \frac{2}{5} + \frac{2}{3}\right) = -\frac{5}{4}
\]
Chuyển vế:
\[
\left(-3 + \frac{3}{x} - \frac{1}{3}\right) = -\frac{5}{4} \cdot \left(1 + \frac{2}{5} + \frac{2}{3}\right)
\]
\[
\left(-3 + \frac{3}{x} - \frac{1}{3}\right) = -\frac{5}{4} \cdot \left(\frac{15}{15} + \frac{6}{15} + \frac{10}{15}\right)
\]
\[
\left(-3 + \frac{3}{x} - \frac{1}{3}\right) = -\frac{5}{4} \cdot \frac{31}{15}
\]
\[
\left(-3 + \frac{3}{x} - \frac{1}{3}\right) = -\frac{31}{12}
\]
\[
-3 + \frac{3}{x} - \frac{1}{3} = -\frac{31}{12}
\]
\[
\frac{3}{x} = -\frac{31}{12} + 3 + \frac{1}{3}
\]
\[
\frac{3}{x} = -\frac{31}{12} + \frac{36}{12} + \frac{4}{12}
\]
\[
\frac{3}{x} = \frac{9}{12}
\]
\[
\frac{3}{x} = \frac{3}{4}
\]
\[
x = 4
\]

c) \(-\frac{3x}{4} \left(\frac{1}{x} + \frac{2}{7}\right) = 0\)

Đầu tiên, ta giải phương trình:
\[
-\frac{3x}{4} \left(\frac{1}{x} + \frac{2}{7}\right) = 0
\]
Vì tích của hai số bằng 0, nên một trong hai số phải bằng 0. Do đó:
\[
-\frac{3x}{4} = 0 \quad \text{hoặc} \quad \left(\frac{1}{x} + \frac{2}{7}\right) = 0
\]
\[
-\frac{3x}{4} = 0 \Rightarrow x = 0 \quad \text{(loại vì x không thể bằng 0 trong phân số)}
\]
\[
\frac{1}{x} + \frac{2}{7} = 0
\]
\[
\frac{1}{x} = -\frac{2}{7}
\]
\[
x = -\frac{7}{2}
\]

d) \(3 - \frac{1 - \frac{1}{2}}{1 + \frac{1}{x}} = \frac{2}{3}\)

Đầu tiên, ta giải phương trình:
\[
3 - \frac{1 - \frac{1}{2}}{1 + \frac{1}{x}} = \frac{2}{3}
\]
Chuyển vế:
\[
3 - \frac{\frac{1}{2}}{1 + \frac{1}{x}} = \frac{2}{3}
\]
\[
3 - \frac{\frac{1}{2}}{\frac{x+1}{x}} = \frac{2}{3}
\]
\[
3 - \frac{\frac{1}{2} \cdot \frac{x}{x+1}} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
Chuyển vế:
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)} = \frac{2}{3}
\]
\[
3 - \frac{x}{2x+2} = \frac{2}{3}
\]
\[
3 - \frac{x}{2(x+1)}