Question

помогите плз алгебра 10-11 класс​

Answer 1

$1)\; \; f(x)=\frac{x-2x^2}{\sqrt{4x-7}}\; \; ,\\\\OJF:\; 4x-7>0\; ,\; \; x>1,75\\\\\boxed {x\in (1,75\, ;+\infty )\; }\\\\\\2)\; \; f(x)=\frac{\sqrt{2-x^2-x}}{x}\\\\OOF:\; \left\{\begin{array}{l}2-x^2-x\geq 0\\x\ne 0\end{array}\right\; \; \left\{\begin{array}{l}x^2+x-2\leq 0\\x\ne 0\end{array}\right\; \; \left\{\begin{array}{l}x\in [-2;1\; ]\\x\ne 0\end{array}\right\; \; \; \Rightarrow \\\\\\\boxed {\; x\in [-2;0\; )\cup (0\, ;1\; ]\; }$

$P.S.\; \; x^2+x-2\leq 0\; ,\; \; (x-1)(x+2)\leq 0\; ,\\\\znaki:\; \; +++[-2\; ]---[\; 1\; ]+++\\\\x\in [-2;1\; ]$

$3)\; \; f(x)=\frac{lg(16-x^2)}{x-2}\\\\OOF:\; \; \left\{\begin{array}{l}16-x^2>0\\x-2\ne 0\end{array}\right\; \; \left\{\begin{array}{ccc}x\in (-4;4)\\x\ne 2\end{array}\right\\\\\\\boxed {\; x\in (-4;2)\cup (2;4)\; }$

$P.S.\; \; 16-x^2>0\; \; \to \; \; x^2-16<0\; \; ,\; \; (x-4)(x+4)<0\; ,\\\\znaki:\; \; +++(-4)---(4)+++\\\\x\in (-4;4\, )$

Answer 2

Ответ: приложено

Объяснение: