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Laster_Alex

Z11

Sep 17th, 2023
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  1. Разложим:
  2. $$\left\{\begin{array}{l}8+2x\ge 0\\ {x}^{2}-6x={\left(2x+8\right)}^{2}\end{array}\right.\Rightarrow \left\{\begin{array}{l} x \ge -4\\ {x}^{2}-6x=4{x}^{2}+32x+64\end{array}\right.$$
  3. Решим квадратное уравнение:
  4. $$3{x}^{2}+38x+64=0\Rightarrow x=\frac{-19\pm \sqrt{{19}^{2}-3\times 64}}{3}=\frac{-19\pm 13}{3}=-\frac{32}{3};-2$$
  5. Под ОДЗ подходит лишь -2
  6.  
  7.  
  8. Ответ:
  9. -2
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