$a\text{)}1-\frac{1}{x}:\left(x:\frac{1}{2x}\right)$  

Założenia:

$x\ne 0\wedge x:\frac{1}{2x}\ne 0$  

$x\ne 0\wedge x\cdot 2x\ne 0$  

$x\ne 0$  

$D=\mathbb{R}-\left\{0\right\}$   

Wykonajmy podane działania:

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

$=1-\frac{1}{x}:2x^2=1-\frac{1}{x}\cdot \frac{1}{2x^2}=1-\frac{1}{2x^3}=$  

$=\frac{2x^3}{2x^3}-\frac{1}{2x^3}=\frac{2x^3-1}{2x^3}$  

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$b\text{)}1-\frac{1}{x}:x:\frac{1}{2x}$  

Założenia:

$x\ne 0\wedge 2x\ne 0$   

$x\ne 0$  

$D=\mathbb{R}-\left\{0\right\}$  

Wykonajmy podane działania:

$1-\frac{1}{x}:x:\frac{1}{2x}=1-\frac{1}{x}\cdot \frac{1}{x}:\frac{1}{2x}=$  

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

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

---

$c\text{)}\left(1-\frac{1}{x}\right):\left(x:\frac{1}{2x}\right)$  

Założenia:

$x\ne 0\wedge x:\frac{1}{2x}\ne 0$  

$x\ne 0\wedge x\cdot 2x\ne 0$  

$x\ne 0$  

$D=\mathbb{R}-\left\{0\right\}$  

Wykonajmy podane działania:

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

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

$=\frac{x-1}{x}\cdot \frac{1}{2x^2}=\frac{x-1}{2x^3}$  

---

$d\text{)}\left(\frac{2}{x}-\frac{x}{2}\right)\cdot \frac{4}{x+2}$  

Założenia:

$x\ne 0\wedge x+2\ne 0$  

$x\ne 0\wedge x\ne -2$   

$D=\mathbb{R}-\left\{-2,0\right\}$  

Wykonajmy podane działania:

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

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

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

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$e\text{)}\left(1+\frac{5}{x-3}\right):\left(2x+4\right)$  

Założenia:

$x-3\ne 0\wedge 2x+4\ne 0$  

$x\ne 3\wedge x\ne -2$   

$D=\mathbb{R}-\left\{-2,3\right\}$  

Wykonajmy podane działania:

$\left(1+\frac{5}{x-3}\right):\left(2x+4\right)=\left(1+\frac{5}{x-3}\right)\cdot \frac{1}{2x+4}=$  

$=\left(\frac{x-3}{x-3}+\frac{5}{x-3}\right)\cdot \frac{1}{2x+4}=\frac{x-3+5}{x-3}\cdot \frac{1}{2x+4}=$  

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

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$f\text{)}\left(\frac{1}{x}-\frac{1}{x+1}\right):\left(\frac{1}{x}+\frac{1}{x+1}\right)$  

Założenia:

$x\ne 0\wedge x+1\ne 0\wedge \frac{1}{x}+\frac{1}{x+1}\ne 0$  

$x\ne 0\wedge x+1\ne 0\wedge \frac{x+1+x}{x\left(x+1\right)}\ne 0$  

$x\ne 0\wedge x+1\ne 0\wedge \frac{2x+1}{x\left(x+1\right)}\ne 0$  

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

$x\ne 0\wedge x\ne -1\wedge x\ne -\frac{1}{2}$  

$D=\mathbb{R}-\left\{-1,-\frac{1}{2},0\right\}$  

Wykonajmy podane działania:

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

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

$=\frac{1}{2x+1}$