a)

$\frac{2{\left(x+2\right)}^3\left(x-5\right)}{{\left(2x+4\right)}^2{\left(x-5\right)}^3}$

Dziedzina:

$2x+4\ne 0\wedge x-5\ne 0$

$x\ne -2x\ne 5$

$x\in \mathbb{R}\setminus \left\{-2,5\right\}$

Uprośćmy to wyrażenie:

$\frac{2{\left(x+2\right)}^3\left(x-5\right)}{{\left(2x+4\right)}^2{\left(x-5\right)}^3}=\frac{2{\left(x+2\right)}^3}{{\left(2\left(x+2\right)\right)}^2{\left(x-5\right)}^2}=\frac{2{\left(x+2\right)}^3}{4{\left(x+2\right)}^2{\left(x-5\right)}^2}=\frac{x+2}{2{\left(x-5\right)}^2}$

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b)

$\frac{x^2-10x+25}{x^2-25}$

Dziedzina:

$x^2-25\ne 0$

$\left(x-5\right)\left(x+5\right)\ne 0$

$x\ne 5\wedge x\ne -5$

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

Uprośćmy to wyrażenie:

$\frac{x^2-10x+25}{x^2-25}=\frac{{\left(x-5\right)}^2}{\left(x+5\right)\left(x-5\right)}=\frac{x-5}{x+5}$

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c)

$\frac{x^3-2x^2}{x^4-4x^2}$

Dziedzina:

$x^4-4x^2\ne 0$

$x^2\left(x^2-4\right)\ne 0$

$x^2\left(x-2\right)\left(x+2\right)\ne 0$

$x\ne 0\wedge x\ne 2\wedge x\ne -2$

$x\in \mathbb{R}\setminus \left\{-2,0,2\right\}$

Uprośćmy to wyrażenie:

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

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d)

$\frac{x^2-6x+9}{x^3-4x^2+3x}$

Dziedzina:

$x^3-4x^2+3x\ne 0$

$x\left(x^2-4x+3\right)\ne 0$

$x\left(x-3\right)\left(x-1\right)\ne 0$

$x\ne 0\wedge x\ne 3\wedge x\ne 1$

$x\in \mathbb{R}\setminus \left\{0,1,3\right\}$

Uprośćmy to wyrażenie:

$\frac{x^2-6x+9}{x^3-4x^2+3x}=\frac{{\left(x-3\right)}^2}{x\left(x-3\right)\left(x-1\right)}=\frac{x-3}{x\left(x-1\right)}$

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e)

$\frac{6x-8}{9x^2-24x+16}$

Dziedzina:

$9x^2-24x+16\ne 0$

${\left(3x-4\right)}^2\ne 0$

$x\ne \frac{4}{3}$

$x\in \mathbb{R}\setminus \left\{\frac{4}{3}\right\}$

Uprośćmy to wyrażenie:

$\frac{6x-8}{9x^2-24x+16}=\frac{6x-8}{{\left(3x-4\right)}^2}=\frac{2\left(3x-4\right)}{{\left(3x-4\right)}^2}=\frac{2}{3x-4}$

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f)

$\frac{3x^2+3x}{2x^3+x^2-x}$

Dziedzina:

$2x^3+x^2-x\ne 0$

$x\left(2x^2+x-1\right)\ne 0$

$x\ne 0\wedge 2x^2+x-1\ne 0$

$\Delta =1^2-4\cdot 2\cdot \left(-1\right)=1+8=9,\sqrt{\Delta }=3$

$x_1\ne \frac{-1-3}{2\cdot 2}=-1$

$x_2\ne \frac{-1+3}{2\cdot 2}=\frac{1}{2}$

$x\in \mathbb{R}\setminus \left\{-1,0,\frac{1}{2}\right\}$

Uprośćmy to wyrażenie:

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