a)

Założenia:

$\left(x-5\right)\left(x+2\right)\ne 0\text{i}\left(x+1\right)\left(x-5\right)\ne 0\text{i}\left(x-3\right)\left(x+2\right)\ne 0$

Wobec tego otrzymujemy:

$x-5\ne 0\text{i}x+2\ne 0\text{i}x+1\ne 0\text{i}x-3\ne 0$

$x\ne 5\text{i}x\ne -2\text{i}x\ne -1\text{i}x\ne 3$

Wykonajmy dzielenie.

$\frac{\left(x-3\right)\left(x+1\right)}{\left(x-5\right)\left(x+2\right)}:\frac{\left(x+1\right)\left(x-5\right)}{\left(x-3\right)\left(x+2\right)}=$

$\frac{\left(x-3\right)\left(x+1\right)}{\left(x-5\right)\left(x+2\right)}\cdot \frac{\left(x-3\right)\left(x+2\right)}{\left(x+1\right)\left(x-5\right)}=$

$=\frac{{\left(x-3\right)}^2}{{\left(x-5\right)}^2}$

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b)

Założenia:

$3x-4\ne 0\text{i}{x}^2+2x\ne 0$

$3x\ne 4\text{i}x\left(x+2\right)\ne 0$

$x\ne \frac{4}{3}\text{i}x\ne 0\text{i}x+2\ne 0$

$x\ne \frac{4}{3}\text{i}x\ne 0\text{i}x\ne -2$

Wykonajmy dzielenie.

$\frac{3x^3-12x}{3x-4}:\left(x^2+2x\right)=$

$=\frac{3x\left(x^2-4\right)}{3x-4}\cdot \frac{1}{x\left(x+2\right)}=$

$=\frac{3x\left(x-2\right)\left(x+2\right)}{3x-4}\cdot \frac{1}{x\left(x+2\right)}=$

$=\frac{3\left(x-2\right)}{3x-4}$

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c)

Założenia:

$9x^2-4\ne 0\text{i}9x-4\ne 0$

$9x^2\ne 4\text{i}9x\ne 4$

$x^2\ne \frac{4}{9}\text{i}x\ne \frac{4}{9}$

$x\ne \frac{2}{3}\text{i}x\ne -\frac{2}{3}\text{i}x\ne \frac{4}{9}$

Wykonajmy dzielenie.

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

$=\frac{x\left(3x-2\right)}{1}\cdot \frac{9x-4}{9x^2-4}=$

$=\frac{x\left(3x-2\right)}{1}\cdot \frac{9x-4}{\left(3x-2\right)\left(3x+2\right)}=$

$=\frac{x\left(9x-4\right)}{3x+2}$

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d)

Założenia:

$x-3\ne 0\text{i}{x}^2-4\ne 0\text{i}{x}^2-2x-3\ne 0$

Rozpatrujemy ostatnie założenie. Wyznaczmy wyróżnik kwadratowy i miejsca zerowe tego trójmianu kwadratowego.

$\Delta ={\left(-2\right)}^2-4\cdot 1\cdot \left(-3\right)=4+12=16$

$x_1=\frac{-\left(-2\right)-4}{2\cdot 1}=\frac{2-4}{2}=\frac{-2}{2}=-1$

$x_2=\frac{-\left(-2\right)+4}{2\cdot 1}=\frac{2+4}{2}=\frac{6}{2}=3$

Zatem otrzymujemy:

$x\ne 3\text{i}{x}^2\ne 4\text{i}x\ne -1\text{i}x\ne 3$

$x\ne 3\text{i}x\ne 2\text{i}x\ne -2\text{i}x\ne -1\text{i}x\ne 3$

$x\ne 3\text{i}x\ne 2\text{i}x\ne -2\text{i}x\ne -1$

Wykonajmy dzielenie.

$\frac{x^2+x-2}{x-3}:\frac{x^2-4}{x^2-2x-3}=$

$=\frac{\left(x+2\right)\left(x-1\right)}{x-3}\cdot \frac{x^2-2x-3}{x^2-4}=$

$=\frac{\left(x+2\right)\left(x-1\right)}{x-3}\cdot \frac{\left(x+1\right)\left(x-3\right)}{\left(x-2\right)\left(x+2\right)}=$

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

$=\frac{x^2-1}{x-2}$

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e)

Założenia:

$3x^2-3\ne 0\text{i}4x+18\ne 0\text{i}x+1\ne 0$

$3x^2\ne 3\text{i}4x\ne -18\text{i}x\ne -1$

$x^2\ne 1\text{i}x\ne -\frac{18}{4}\text{i}x\ne -1$

$x\ne 1\text{i}x\ne -1\text{i}x\ne -\frac{9}{2}$

Wykonajmy dzielenie.

$\frac{4x^2-81}{3x^2-3}:\frac{4x+18}{x+1}=$

$=\frac{\left(2x-9\right)\left(2x+9\right)}{3\left(x^2-1\right)}\cdot \frac{x+1}{4x+18}=$

$=\frac{\left(2x-9\right)\left(2x+9\right)}{3\left(x-1\right)\left(x+1\right)}\cdot \frac{x+1}{2\left(2x+9\right)}=$

$\frac{2x-9}{3\left(x-1\right)}\cdot \frac{1}{2}=\frac{2x-9}{6\left(x-1\right)}=\frac{2x-9}{6x-6}$

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f)

Założenia:

${\left(2x-6\right)}^2\ne 0\text{i}3x^3+27x\ne 0\text{i}4x+12\ne 0$

$2x-6\ne 0\text{i}3x\left(x^2+9\right)\ne 0\text{i}4\left(x+3\right)\ne 0$

$2\left(x-3\right)\ne 0\text{i}3x\ne 0\text{i}{x}^2+9\ne 0\text{i}x+3\ne 0$

$x-3\ne 0\text{i}x\ne 0\text{i}{x}^2\ne -9\text{i}x\ne -3$

$x\ne 3\text{i}x\ne 0\text{i}x\in \mathbb{R}\text{i}x\ne -3$

Wykonajmy dzielenie.

$\frac{x^4-81}{{\left(2x-6\right)}^2}:\frac{3x^3+27x}{4x+12}=$

$=\frac{\left(x^2-9\right)\left(x^2+9\right)}{{\left(2\left(x-3\right)\right)}^2}\cdot \frac{4x+12}{3x^3+27x}=$

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

$=\frac{{\left(x+3\right)}^2}{\left(x-3\right)\cdot 3x}=\frac{{\left(x+3\right)}^2}{3x\left(x-3\right)}$