a)

Najpierw dziedzina:

$$\begin{gathered} x^2+2x+1\ne 0\text{i}x-5\ne 0\text{i}x+1\ne 0 \\ (x+1{)}^2\ne 0\text{i}x\ne 5\text{i}x\ne -1 \\ x+1\ne 0 \\ x\ne -1 \end{gathered}$$

$D=\mathbb{R}\setminus \left\{-1,5\right\}$

Wykonajmy to dzielenie:

$\frac{x^2-25}{x^2+2x+1}:\frac{x-5}{x+1}=\frac{\cancel{\left(x-5\right)}\left(x+5\right)}{{\left(x+1\right)}^{\cancel{2}}}\cdot \frac{\cancel{x+1}}{\cancel{x-5}}=\frac{x+5}{x+1}$

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b)

Najpierw dziedzina:

$$\begin{gathered} x-1\ne 0\text{i}{x}^2+4x+4\ne 0\text{i}{x}^2-1\ne 0 \\ x\ne 1\text{i}(x+2{)}^2\ne 0\text{i}(x-1)(x+1)\ne 0 \\ x\ne -2\text{i}x\ne 1\text{i}x\ne -1 \end{gathered}$$

$D=\mathbb{R}\setminus \left\{-2,-1,1\right\}$

Wykonajmy to dzielenie:

$\frac{x^2-4}{x-1}:\frac{x^2+4x+4}{x^2-1}=\frac{\left(x-2\right)\cancel{\left(x+2\right)}}{\cancel{x-1}}\cdot \frac{\cancel{\left(x-1\right)}\left(x+1\right)}{{\left(x+2\right)}^{\cancel{2}}}=\frac{\left(x-2\right)\left(x+1\right)}{x+2}$

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c)

Najpierw dziedzina:

$$\begin{gathered} x^2-6x+9\ne 0\text{i}{x}^2+4x\ne 0\text{i}{x}^2-9\ne 0 \\ (x-3{)}^2\ne 0\text{i}x(x+4)\ne 0\text{i}(x-3)(x+3)\ne 0 \\ x\ne 3\text{i}x\ne 0\text{i}x\ne -4\text{i}x\ne 3\text{i}x\ne -3 \end{gathered}$$

$D=\mathbb{R}\setminus \left\{-4,-3,0,3\right\}$

Wykonajmy to dzielenie:

$\frac{x+4}{x^2-6x+9}:\frac{x^2+4x}{x^2-9}=\frac{\cancel{x+4}}{{\left(x-3\right)}^{\cancel{2}}}\cdot \frac{\cancel{\left(x-3\right)}\left(x+3\right)}{x\cancel{\left(x+4\right)}}=\frac{x+3}{x\left(x-3\right)}$