2.1.4) \( \frac{3^{n-1} \cdot 9^{n-3}}{27^{n}} \) 3.1.1) \( \frac{4 y-8 y^{2}}{4 y^{3}} \) (2) \( \frac{y^{2}+6 y+7}{y^{3}-27} \) 3) \( \frac{x^{2}-x}{2 x+2} \times \frac{4 x}{3 x-3} \)

**2.1.4)** Simplify \[ \frac{3^{n-1} \cdot 9^{n-3}}{27^n}. \] 1. Express \(9\) and \(27\) as powers of \(3\): - \(9=3^2\) - \(27=3^3\) 2. Rewrite the expression: \[ \frac{3^{n-1} \cdot (3^2)^{n-3}}{(3^3)^n} = \frac{3^{n-1} \cdot 3^{2(n-3)}}{3^{3n}}. \] 3. Simplify the exponents in the numerator: \[ 3^{n-1} \cdot 3^{2n-6} = 3^{(n-1+2n-6)} = 3^{3n-7}. \] 4. Write the fraction as a single power of \(3\): \[ \frac{3^{3n-7}}{3^{3n}} = 3^{(3n-7-3n)} = 3^{-7}. \] Thus, the simplified result is: \[ 3^{-7}. \] --- **3.1.1)** Simplify \[ \frac{4y-8y^2}{4y^3}. \] 1. Factor the numerator: \[ 4y-8y^2 = 4y(1-2y). \] 2. Factor the denominator: \[ 4y^3 = 4y^3. \] 3. Cancel the common factor \(4y\) (with \(y \neq 0\)): \[ \frac{4y(1-2y)}{4y^3} = \frac{1-2y}{y^2}. \] Thus, the simplified result is: \[ \frac{1-2y}{y^2}. \] --- **(2)** Simplify \[ \frac{y^{2}+6y+7}{y^{3}-27}. \] 1. Notice the denominator is a difference of cubes: \[ y^3-27 = y^3-3^3 = (y-3)(y^2+3y+9). \] 2. The numerator \(y^2+6y+7\) does not factor nicely (its discriminant is \(36-28=8\)), so it remains as is. Thus, the expression in factored form is: \[ \frac{y^2+6y+7}{(y-3)(y^2+3y+9)}. \] --- **3)** Simplify \[ \frac{x^{2}-x}{2x+2} \times \frac{4x}{3x-3}. \] 1. Factor each expression: - Numerator of first fraction: \[ x^2-x = x(x-1). \] - Denominator of first fraction: \[ 2x+2 = 2(x+1). \] - Denominator of second fraction: \[ 3x-3 = 3(x-1). \] 2. Rewrite the product: \[ \frac{x(x-1)}{2(x+1)} \times \frac{4x}{3(x-1)}. \] 3. Cancel the common factor \((x-1)\) (with \(x \neq 1\)): \[ \frac{x}{2(x+1)} \times \frac{4x}{3}. \] 4. Multiply the remaining factors: \[ \frac{4x^2}{6(x+1)}. \] 5. Simplify by dividing numerator and denominator by 2: \[ \frac{2x^2}{3(x+1)}. \] Thus, the simplified result is: \[ \frac{2x^2}{3(x+1)}. \]