Question

Помогитеее пожалуйста с решением прошу!!!!

Answer 1

Ответ:  Б .

$24)\ \ 1.\ \ \ f(x)=\sqrt{6x-2}+\sqrt{3-x}\ \ ,\\\\OOF:\ \left\{\begin{array}{l}6x-2\geq 0\\3-x\geq 0\end{array}\right\ \ \left\{\begin{array}{l}x\geq \dfrac{1}{3}\\x\leq 3\end{array}\right\ \ \ \Rightarrow \ \ \ x\in \Big[\ \dfrac{1}{3}\ ;\ 3\ \Big]$

$2.\ \ \ f(x)=\dfrac{7x-1}{\sqrt{6x-2}}-\sqrt{3-x}\ \ ,\\\\OOF:\ \left\{\begin{array}{l}6x-2>0\\3-x\geq 0\end{array}\right\ \ \left\{\begin{array}{l}x>\dfrac{1}{3}\\x\leq 3\end{array}\right\ \ \ \Rightarrow \ \ \ \boxed{\ \ x\in \Big(\ \dfrac{1}{3}\ ;\ 3\ \Big]\ \ }$

$3.\ \ \ f(x)=\dfrac{x}{\sqrt{3-x}}-\dfrac{1}{\sqrt{6x-2}}\ \ ,\\\\OOF:\ \left\{\begin{array}{l}6x-2>0\\3-x>0\end{array}\right\ \ \left\{\begin{array}{l}x>\dfrac{1}{3}\\x>3\end{array}\right\ \ \ \Rightarrow \ \ \ x\in \Big(\ \dfrac{1}{3}\ ;\ 3\ \Big)$

$4.\ \ \ f(x)=\sqrt{6x-2}+\dfrac{x+3}{\sqrt{3-x}}\ \ ,\\\\OOF:\ \left\{\begin{array}{l}6x-2\geq 0\\3-x>0\end{array}\right\ \ \left\{\begin{array}{l}x\geq \dfrac{1}{3}\\x>3\end{array}\right\ \ \ \Rightarrow \ \ \ x\in \Big[\ \dfrac{1}{3}\ ;\ 3\ \Big)$