Simplify each of the following: \( 2.1(2 a-3)(3 a+1) \) \( 2.2(x+3)\left(x^{2}-3 x+4\right) \) \( 2.3\left(\frac{2}{x}+\frac{x}{2}\right)^{2} \)

\[ \textbf{Problem 2.1:} \quad 2.1(2a-3)(3a+1) \] First, expand the product \((2a-3)(3a+1)\): \[ \begin{aligned} (2a-3)(3a+1) &= 2a \cdot 3a + 2a \cdot 1 - 3 \cdot 3a - 3 \cdot 1\\[1mm] &= 6a^2 + 2a - 9a - 3\\[1mm] &= 6a^2 - 7a - 3. \end{aligned} \] Multiply by \(2.1\): \[ 2.1(6a^2 - 7a - 3) = 12.6a^2 - 14.7a - 6.3. \] --- \[ \textbf{Problem 2.2:} \quad 2.2(x+3)(x^2-3x+4) \] First, expand the product \((x+3)(x^2-3x+4)\): \[ \begin{aligned} (x+3)(x^2-3x+4) &= x(x^2-3x+4) + 3(x^2-3x+4)\\[1mm] &= x^3 - 3x^2 + 4x + 3x^2 - 9x + 12\\[1mm] &= x^3 + \underline{(-3x^2+3x^2)} -5x + 12\\[1mm] &= x^3 - 5x + 12. \end{aligned} \] Multiply by \(2.2\): \[ 2.2(x^3-5x+12) = 2.2x^3 - 11x + 26.4. \] --- \[ \textbf{Problem 2.3:} \quad 2.3\left(\frac{2}{x}+\frac{x}{2}\right)^{2} \] First, simplify the expression inside the parentheses: \[ \frac{2}{x}+\frac{x}{2} \] Find a common denominator (which is \(2x\)): \[ \frac{2}{x} = \frac{4}{2x}, \quad \frac{x}{2} = \frac{x^2}{2x}. \] Thus, \[ \frac{2}{x}+\frac{x}{2} = \frac{4+x^2}{2x}. \] Now square the expression: \[ \left(\frac{4+x^2}{2x}\right)^2 = \frac{(4+x^2)^2}{4x^2}. \] Multiply by \(2.3\): \[ 2.3\left(\frac{(4+x^2)^2}{4x^2}\right) = \frac{2.3}{4}\cdot\frac{(4+x^2)^2}{x^2}. \] Note that \(\frac{2.3}{4}\) can be written as \(0.575\) or, using fractions, note \(2.3=\frac{23}{10}\) so that \[ \frac{2.3}{4} = \frac{23}{10}\cdot\frac{1}{4} = \frac{23}{40}. \] Thus, the simplified expression is: \[ \frac{23}{40}\cdot\frac{(4+x^2)^2}{x^2}. \] --- \[ \boxed{ \begin{aligned} 2.1(2a-3)(3a+1) &= 12.6a^2 - 14.7a - 6.3,\\[1mm] 2.2(x+3)(x^2-3x+4) &= 2.2x^3 - 11x + 26.4,\\[1mm] 2.3\left(\frac{2}{x}+\frac{x}{2}\right)^2 &= \frac{23}{40}\cdot\frac{(4+x^2)^2}{x^2}. \end{aligned} } \]