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- 1)
- По УРК5 $$\sqrt{x+7}\ge 0.5(x-1)⟺\left[\begin{array}{l}\left\{\begin{array}{l}0.5(x-1)<0\\ x+7\ge 0\end{array}\right.\\ \left\{\begin{array}{l}0.5(x-1)\ge 0\\ x+7\ge 0.25({x}^{2}-2x+1)\end{array}\right.\end{array}\right.⟺\left[\begin{array}{l}\left\{\begin{array}{l}x<1\\ x\ge -7\end{array}\right.\\ \left\{\begin{array}{l}x\ge 1\\ 4x+28\ge {x}^{2}-2x+1\end{array}\right.\end{array}\right.⟺\left[\begin{array}{l}x\in \left[-7;1\right)\\ \left\{\begin{array}{l}x\ge 1\\ {x}^{2}-6x-27\le 0\end{array}\right.\end{array}\right.$$
- 2)
- Решим квадратное уравнение:
- $$x=3\pm \sqrt{9+27}=3\pm 6=-3;9$$
- Т.к. старший коэфф. >0, $$x\in \left[-3;9\right]$$
- 3)
- Выходит:
- $$\left[\begin{array}{l}x\in \left[-7;1\right)\\ \left\{\begin{array}{l}x\ge 1\\ x\in \left[-3;9\right]\end{array}\right.\end{array}\right.\Rightarrow x\in \left[-7;9\right]$$
- Длина данного промежутка 16
- Ответ:
- 16
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