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[Гост]

4([tex]x^{2 }[/tex]-9)/2[tex]x^{2 }[/tex] +x-15

[ammornil]

4([tex]x^{2 }[/tex]-9)/2[tex]x^{2 }[/tex] +x-15

$\\[12pt]$ $a^{2}-b^{2}= (a-b)\cdot{}(a+b), \quad x^{2}-9= x^{2} -3^{2}= (x-3)(x+3) \\[6pt] ax^{2} +bx +c= a\cdot{}(x-x_{1})\cdot{}(x-x_{2}), \quad x_{1,2}=\dfrac{-b\pm\sqrt{b^{2}-4\cdot{a}\cdot{c}}}{2\cdot{a}} \Rightarrow \\[6pt] \hspace{5em} 2x^{2} +1x -15 \rightarrow x_{1,2}= \dfrac{-1\pm\sqrt{1^{2}-4\cdot{2}\cdot{(-15)}}}{2\cdot{2}}=\dfrac{-1\pm 11}{4} \Rightarrow \begin{cases} x_{1}=-3 \\ x_{2}=\dfrac{5}{2} \end{cases} \\[6pt] \hspace{5em} x^{2} +x -15= 2(x+3)\left(x-\dfrac{5}{2} \right) \\[12pt] A= \dfrac{4(x^{2}-9)}{2x^{2} +x -15}= \dfrac{4(x-3)\cancel{(x+3)}}{2\cancel{(x+3)}\left(x-\dfrac{5}{2} \right)}= \frac{4x -12}{2x -5}$

[Гост]

Благодаря!