nathi123 » Пт июл 07, 2017 8:07 pm
4. зад. Пусть х - катеты прямоуг.равноб. треугольника [tex]\Rightarrow c=x\sqrt{2}[/tex] гипотенуза (теорема Пифагора);
[tex]p=\frac{1}{2} ( x+x+x\sqrt{2})=\frac{x(2+\sqrt{2}}{2}; r = \sqrt{2}=p-c ; p-c=\frac{x(2-\sqrt{2})}{2}\Rightarrow \sqrt{2}=\frac{x(2-\sqrt{2)}}{2}[/tex]
[tex]x=\frac{2\sqrt{2}}{(2-\sqrt{2})}\Rightarrow p=\frac{2\sqrt{2}(2+\sqrt{2})}{(2-\sqrt{2}}=\sqrt{2}(2+\sqrt{2})^{2}[/tex]
[tex]S=pr=\sqrt{2}(2+\sqrt{2})^{2}.\sqrt{2}=2(4+2+4\sqrt{2})=12+8\sqrt{2}[/tex].
6.зад. [tex](\sqrt{x})^{2} + \sqrt{(x^{2} - 9)} = x +4\Leftrightarrow\begin{array}{|l} x\ge 0\\ x^{2}- 9 \ge 0\\ x+4\ge 0 \\ x+\sqrt{(x^{2}-9)}=x+4\end{array}[/tex]
[tex]\begin{array}{|l} x \in[3,+\infty) \\ x^{2}-9=16 \end{array} \Leftrightarrow \begin{array}{|l} x\in [3,+\infty)\\ x^{2} =5\end{array}\Leftrightarrow x=5[/tex].