Question

Зробіть, будь ласка, 7,8,9 завдання. Даю 30 балів! Терміново!​

Answer 1

Ответ:

$\displaystyle 7)\ \ \frac{18}{3+\sqrt3}=\frac{3\cdot 6}{\sqrt3\cdot (\sqrt3+1)}=\frac{6\sqrt3}{\sqrt3+1}=\frac{6\sqrt3\cdot (\sqrt3-1)}{(\sqrt3+1)(\sqrt3-1)}=\\\\\\=\frac{6\sqrt3\cdot (\sqrt3-1)}{3-1}=\frac{6\sqrt3\cdot (\sqrt3-1)}{2}=2\sqrt3\cdot (\sqrt3-1)=6-2\sqrt3\\\\\\8)\ \ f(x)=\frac{1}{\sqrt{x}-1}+\frac{1}{x^2-2x}\ \ \ \ \Rightarrow$

$\left\{\begin{array}{lll}x\geq 0\\\sqrt{x}-1\ne 0\\x^2-2x\ne 0\end{array}\right\ \ \left\{\begin{array}{lll}x\geq 0\\\sqrt{x}\ne 1\\x(x-2)\ne 0\end{array}\right\ \ \left\{\begin{array}{lll}x\geq 0\\x\ne 1\\x\ne 0\ ,\ x\ne 2\end{array}\right\ \ \ \ \Rightarrow \\\\\\x\in (\ 0\ ;\ 1\ )\cup (\ 1\ ;\ 2\ )\cup (\ 2\ ;+\infty \, )$

$9)\ \ f(x)=|x|+3\\\\|x|\geq 0\ \ \ \Rightarrow \ \ \ |x|+3\geq 3\ \ \ \Rightarrow \ \ \ f(x)\in [\ 3\ ;+\infty \, )$