Odpowiedź:
[tex]\displaystyle \frac{2x^{2} +5x-3}{9-x^{2} }\quad x\in R-\{-3,3\} \\\Delta =25+24=49\quad \sqrt{\Delta} =7\\x_1=\frac{-5+7}{4} =\frac{2}{4} =\frac{1}{2} \qquad x_2=\frac{-5-7}{4} =\frac{-12}{4} =-3\\2x^{2} +5x-3=2(x-\frac{1}{2} )(x+3)=(2x-1)(x+3)\\\frac{2x^{2} +5x-3}{9-x^{2} }=\frac{(2x-1)(x+3)}{(3-x)(3+x)} =\frac{2x-1}{3-x} =\frac{1-2x}{x-3}[/tex]
C
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Odpowiedź:
[tex]\displaystyle \frac{2x^{2} +5x-3}{9-x^{2} }\quad x\in R-\{-3,3\} \\\Delta =25+24=49\quad \sqrt{\Delta} =7\\x_1=\frac{-5+7}{4} =\frac{2}{4} =\frac{1}{2} \qquad x_2=\frac{-5-7}{4} =\frac{-12}{4} =-3\\2x^{2} +5x-3=2(x-\frac{1}{2} )(x+3)=(2x-1)(x+3)\\\frac{2x^{2} +5x-3}{9-x^{2} }=\frac{(2x-1)(x+3)}{(3-x)(3+x)} =\frac{2x-1}{3-x} =\frac{1-2x}{x-3}[/tex]
C