Решить неравенства. заранее большое ! 1) (1/4)^(x-x^2)> 1/2 2) (корень из (2x^3-22x^2+60x))/(x-6)> =(2x-10)

Показать ответ

Ответ:

rih904507

06.10.2020 20:01

$1)\; \; \Big (\frac{1}{4}\Big )^{x-x^2}\ \textgreater \ \frac{1}{2}\\\\\Big ( \frac{1}{2}\Big )^{2(x-x^2)}\ \textgreater \ \frac{1}{2}\quad \Rightarrow \quad 2(x-x^2)0\\\\ D/4=1-2=-1\ \textless \ 0\; \; \Rightarrow \; \; net\; kornej\; i\; \; tak\; kak\; \; a=2\ \textgreater \ 0\; ,\; to\\\\2x^2-2x+1\ \textgreater \ 0\; pri\; x\in R\\\\Otvet:\; \; x\in (-\infty ,+\infty )\; .$

$2)\; \; \frac{\sqrt{2x^3-22x^2+60x}}{x-6} \geq 2x-10\; ,\\\\ODZ:\; \; \left \{ {{2x^3-22x^2+60x \geq 0} \atop {x-6\ne 0}} \right. \; \; \left \{ {{2x(x^2-11x+30) \geq 0} \atop {x\ne 6}} \right. \\\\x^2-11x+30=0\; ,\; \; D=1\; ,\; \; x_1=5\; ,\; \; x_2=6\; \; \Rightarrow \\\\2x(x-5)(x-6) \geq 0\; ,\; \; x\ne 6\\\\---[\, 0\, ]+++[\, 5\, ]---( 6)+++\\\\\underline {x\in [\, 0,5\, ]\cup (6,+\infty )}\\\\\\ \frac{\sqrt{2x(x-5)(x-6)}}{x-6}-2(x-5) \geq 0 \\\\ \frac{\sqrt{2x(x-5)(x-6)}-2(x-5)(x-6)}{x-6} \geq 0$

$\frac{\sqrt{2(x-5)(x-6)}\cdot \Big (\sqrt{x}-\sqrt{2(x-5)(x-6)}\Big )}{x-6} \geq 0\\\\Tak\; kak\; \; \sqrt{2(x-5)(x-6) }\geq 0\; ,\; \; to\; \; \; \frac{\sqrt{x}-\sqrt{2(x-5)(x-6)}}{x-6} \geq 0\\\\a)\; \; \left \{ {{\sqrt{x}-\sqrt{2(x-5)(x-6)} \geq 0} \atop {x-6\ \textgreater \ 0}} \right. \quad ili\quad b)\; \; \left \{ {{\sqrt{x}-\sqrt{2(x-5)(x-6)} \leq 0} \atop {x-6\ \textless \ 0}} \right. \\\\a)\; \; x\ \textgreater \ 6\; \; \; i\; \; \; \sqrt{2(x-5)(x-6)} \leq \sqrt{x}\; \; \Leftrightarrow \left \{ {{2(x-5)(x-6)\geq 0} \atop {2(x-5)(x-6)\leq x} \right.$

$\left \{ {{x\in [\, 0,5\, ]\cup (6,+\infty )} \atop {2x^2-22x+60-x \leq 0}} \right. \; \; \Rightarrow \; \underline {x\in [\, 4;5\, ]\cup (6;\; 7,5\, ]}\; \; ,\; \; tak\; kak\\\\2x^2-23x+60 \leq 0\; ,\; \; D=49\; ,\; x_1=4\; ,\; \; x_2=7,5\\\\+++[\, 4\, ]---[7,5\, ]+++\quad x\in [\, 4;\; 7,5\, ]\\\\ \left \{ {{x\ \textgreater \ 6} \atop {x\in [\, 4;5\, ]\cup (6;\; 7,5\, ]}} \right. \; \; \to \; \; \; \underline {x\in (6;\; 7,5\, ]}$

$b)\; \; x\ \textless \ 6\; \; i\; \; \; \sqrt{x} \leq \sqrt{2(x-5)(x-6)}\; \; \Leftrightarrow \; \; \left \{ {{x \geq 0} \atop {x \leq 2(x-5)(x-6)}} \right.$

$\left \{ {{0\leq x\ \textless \ 6} \atop {2x^2-23x+60 \geq 0}} \right. \; \; \left \{ {{0\leq x\ \textless \ 6} \atop {x\in (-\infty ,4\, ]\cup [\, 7,5\; ;+\infty )}} \right. \; \; \Rightarrow \; \; x\in [\, 0,4\, ]\\\\ \left \{ {{x\in [\, 0,5\, ]\cup (6,+\infty )} \atop {x\in [\, 0,4\, ]}} \right. \; \; \Rightarrow \; \; \; \underline {x\in [\, 0,4\, ]\cup \{5\}}\\\\Otvet:\; \; x\in [\, 0;\, 4\, ]\cup \{5\}\cup(6;\; 7,5\, ]\; .$