Odpowiedź:
Szczegółowe wyjaśnienie:
[tex]\displaystyle \frac{x^{2}-4 }{3-x} \cdot\frac{x-3}{x+2} =\frac{(x-2)(x+2)}{-(x-3)} \cdot\frac{x-3}{x+2} =\frac{x-2}{-1} =2-x\quad x\in R-\{-2,3\}\\b)\\\frac{-x^2+2x}{x+1} :\frac{2x-4}{x^{2} -1} =\frac{-x(x-2)}{x+1} \cdot\frac{(x-1)(x+1)}{2(x-2)} =\frac{-x(x-1)}{2} \\\quad x\in R-\{-1,1,2\}[/tex]
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Odpowiedź:
Szczegółowe wyjaśnienie:
[tex]\displaystyle \frac{x^{2}-4 }{3-x} \cdot\frac{x-3}{x+2} =\frac{(x-2)(x+2)}{-(x-3)} \cdot\frac{x-3}{x+2} =\frac{x-2}{-1} =2-x\quad x\in R-\{-2,3\}\\b)\\\frac{-x^2+2x}{x+1} :\frac{2x-4}{x^{2} -1} =\frac{-x(x-2)}{x+1} \cdot\frac{(x-1)(x+1)}{2(x-2)} =\frac{-x(x-1)}{2} \\\quad x\in R-\{-1,1,2\}[/tex]