1. **Evaluate the expression without a calculator:** Given expression: $$\left(\frac{1}{2} + \frac{1}{2} - \frac{1}{2} \right) \div \left(\frac{1}{3} + \frac{1}{7} \times \frac{6}{5} \times \frac{3}{2} \right)$$ Step 1: Simplify inside the first parentheses: $$\frac{1}{2} + \frac{1}{2} - \frac{1}{2} = \frac{1}{2}$$ Step 2: Simplify inside the second parentheses: Calculate the multiplication first: $$\frac{1}{7} \times \frac{6}{5} = \frac{6}{35}$$ Then multiply by $$\frac{3}{2}$$: $$\frac{6}{35} \times \frac{3}{2} = \frac{18}{70} = \frac{9}{35}$$ Now add $$\frac{1}{3}$$: $$\frac{1}{3} + \frac{9}{35} = \frac{35}{105} + \frac{27}{105} = \frac{62}{105}$$ Step 3: Divide the results: $$\frac{1}{2} \div \frac{62}{105} = \frac{1}{2} \times \frac{105}{62} = \frac{105}{124}$$ 2. **Simplify by factorization:** Expression: $$\frac{x^2 - xy + y^2}{xy - x - y} - \frac{2x^2 - 5xy + 3y^2}{3xy - 2x - y}$$ Step 1: Factor denominators: - For $$xy - x - y$$, factor by grouping: $$xy - x - y = x(y - 1) - y = (x - 1)(y - 1)$$ - For $$3xy - 2x - y$$, rewrite as: $$3xy - 2x - y = x(3y - 2) - y$$ No simple factorization; keep as is. Step 2: Factor numerators if possible: - $$x^2 - xy + y^2$$ does not factor nicely over integers. - $$2x^2 - 5xy + 3y^2$$ factors as: $$(2x - 3y)(x - y)$$ Step 3: Rewrite expression: $$\frac{x^2 - xy + y^2}{(x - 1)(y - 1)} - \frac{(2x - 3y)(x - y)}{3xy - 2x - y}$$ Since denominators differ and no further simplification is straightforward, this is the simplified form. 3. **Security light poles problem:** Given: - Poles on left side are $$a$$ meters apart. - Poles on right side are $$b$$ meters apart. - Poles are directly opposite at one end. Step 1: The poles will be directly opposite again when the distances from the start are equal: $$na = mb$$ where $$n$$ and $$m$$ are integers representing the number of poles on each side. Step 2: The first time this happens (other than zero) is at the least common multiple (LCM) of $$a$$ and $$b$$: $$\text{LCM}(a,b) = na = mb$$ Step 3: Number of poles on each side: $$n = \frac{\text{LCM}(a,b)}{a} + 1$$ (including the first pole at zero) $$m = \frac{\text{LCM}(a,b)}{b} + 1$$ Step 4: Total poles erected: $$n + m$$ **Final answers:** 1. $$\frac{105}{124}$$ 2. Simplified expression as above. 3. Total poles $$= \frac{\text{LCM}(a,b)}{a} + \frac{\text{LCM}(a,b)}{b} + 2$$