gimplify bhe following \[ \begin{array}{l} \frac{1 a d x}{3}-\frac{4 x}{12} \\ \text { 2. } \frac{0 b-a^{2}}{a^{2}+2 a b+b^{2}} \div \frac{a^{2}}{a^{2}-b^{2}} \times \frac{a^{2}+a b}{a^{2}-2 a b+b^{2}} \\ 3 \cdot \frac{4}{4}-\frac{2}{4^{2}} \\ 4 \cdot \frac{3}{x-9}-\frac{2}{x+3} \\ 5 \cdot \frac{3}{x^{2}+6 x+9}-\frac{2}{x^{2}-9}-\frac{1}{x^{2}-6 x+9} \end{array} \]

Let's simplify these expressions one by one in a playful and engaging way! 1. \(\frac{1 a d x}{3} - \frac{4 x}{12}\): We can simplify \(\frac{4 x}{12}\) to \(\frac{x}{3}\) because \(4 \div 12 = \frac{1}{3}\). Thus, the expression becomes: \[ \frac{1 a d x}{3} - \frac{x}{3} = \frac{(a d - 1)x}{3}. \] 2. \(\frac{0 b - a^{2}}{a^{2}+2ab+b^{2}} \div \frac{a^{2}}{a^{2}-b^{2}} \times \frac{a^{2}+ab}{a^{2}-2ab+b^{2}}\): The numerator simplifies as \(0b - a^{2} = -a^2\), and the denominator \(a^{2}+2ab+b^{2} = (a+b)^{2}\). This gives us: \[ -\frac{a^{2}}{(a+b)^{2}} \div \frac{a^{2}}{(a-b)(a+b)} \times \frac{(a+b)(a)}{(a-b)^{2}}. \] When we divide, we multiply by the reciprocal. Let’s cancel \(a^2\) and simplified terms to get: \[ = -\frac{(a+b)(a)}{(a-b)(a+b)} = -\frac{a}{a-b}, \quad \text{as long as } a \neq -b. \] 3. \(3 \cdot \frac{4}{4} - \frac{2}{4^{2}}\): Since \(3 \cdot \frac{4}{4} = 3\), and \(\frac{2}{4^2} = \frac{2}{16} = \frac{1}{8}\), we have: \[ 3 - \frac{1}{8} = \frac{24}{8} - \frac{1}{8} = \frac{23}{8}. \] 4. \(4 \cdot \frac{3}{x-9} - \frac{2}{x+3}\): Here, we multiply \(4 \cdot \frac{3}{x-9}\) to get \(\frac{12}{x-9}\). Then, let's express \(\frac{2}{x+3}\) over a common denominator of \((x-9)(x+3)\): \[ \frac{12(x+3) - 2(x-9)}{(x-9)(x+3)} = \frac{12x + 36 - 2x + 18}{(x-9)(x+3)} = \frac{10x + 54}{(x-9)(x+3)}. \] 5. \(5 \cdot \frac{3}{x^{2}+6x+9} - \frac{2}{x^{2}-9} - \frac{1}{x^{2}-6x+9}\): Simplifying this, we know \(x^2 + 6x + 9 = (x+3)^2\), \(x^2 - 9 = (x-3)(x+3)\), and \(x^2 - 6x + 9 = (x-3)^2\): \[ 5 \cdot \frac{3}{(x + 3)^2} - \frac{2}{(x-3)(x+3)} - \frac{1}{(x-3)^2}. \] The common denominator becomes \((x-3)^2(x + 3)^2\). After combining, you will get an expression with polynomial coefficients which can be handled similarly through algebraic manipulation. And there you have it! Each piece of the puzzle simplified to showcase a clearer form. Happy calculating!