[tex]\frac{2x^{2}+5x-3}{10x-20}:\frac{4x^{2}-1}{3x-6}[/tex]
Założenia:
[tex]10x-20 \neq 0\\x-2 \neq 0 \ \ \Rightarrow \ \ x\neq 2\\i\\3x-6 \neq 0\\x-2 \neq 0 \ \ \Rightarrow \ \ x \neq 2\\i\\4x^{2}-1\neq 0\\(2x+1)(2x-1) \neq 0 \ \ \rightarrow \ \ x \neq -\frac{1}{2} \ \wedge \ x \neq \frac{1}{2}\\x \in R\setminus\{-\frac{1}{2}, \frac{1}{2}, 2\}[/tex]
[tex]2x^{2}+5x-3 = 2x^{2}+6x-x-3 =2x(x+3)-(x+3) = (x+3)(2x-1)[/tex]
[tex]\dfrac{2x^{2}+5x-3}{10x-20}:\dfrac{4x^{2}-1}{3x-6} = \dfrac{(x+3)(2x-1)}{10(x-2)}\cdot\dfrac{3(x-2)}{(2x-1)(2x+1)}=\dfrac{3(x+3)}{10(2x+1)}=\\\\\\=\dfrac{3x+9}{20x+10}[/tex]
