Решить рациональное неравенство

Question 1

Question 2

$5x^2-14x-3-\frac{1}{3+14x-5x^2}\leq -2\; ,\\\\ODZ:\; 5x^2-14x-3\ne 0\; ,\; \; x\ne -0,2\; ,\; \; x\ne 3\\\\zamena:\; \; t=5x^2-14x-3\; \; \to \; \; \; t-\frac{1}{-t}\leq -2\; \; ,\; \; t+\frac{1}{t}+2\leq 0\; ,\\\\\frac{t^2+2t+1}{t}\leq 0\; ,\; \; \frac{(t+1)^2}{t}\leq 0\; \; \Rightarrow \; \; t\in \{-1\}\cup (0,+\infty )\; ,\\\\t=-1\; ,\; \; 5x^2-14x-3=-1\; ,\; \; 5x^2-14x-2=0\; ,\\\\D/4=7^2+5\cdot 2=59\; ,\; \; x_{1,2}=\frac{7\pm \sqrt{59}}{5}\\\\x_1\approx -0,136\; \; ,\; \; x_2\approx 2,936$

0\; \; \to \; \; 5x^2-14x-3>0\; ,\; x\in (-\infty ;\, -0,2)\cup (3,+\infty )\\\\Otvet:\; \; x\in (-\infty ;\, -0,2)\cup \{\frac{7-\sqrt{59}}{5}\}\cup\{\frac{7+\sqrt{59}}{5}\}\cup (3,+\infty )\; ." alt="t>0\; \; \to \; \; 5x^2-14x-3>0\; ,\; x\in (-\infty ;\, -0,2)\cup (3,+\infty )\\\\Otvet:\; \; x\in (-\infty ;\, -0,2)\cup \{\frac{7-\sqrt{59}}{5}\}\cup\{\frac{7+\sqrt{59}}{5}\}\cup (3,+\infty )\; ." align="absmiddle" class="latex-formula">

NNNLLL54_zn · Answer 1 · 2020-06-01T17:11:59+0000

$5x^2-14x-3-\frac{1}{3+14x-5x^2}\leq -2\; ,\\\\ODZ:\; 5x^2-14x-3\ne 0\; ,\; \; x\ne -0,2\; ,\; \; x\ne 3\\\\zamena:\; \; t=5x^2-14x-3\; \; \to \; \; \; t-\frac{1}{-t}\leq -2\; \; ,\; \; t+\frac{1}{t}+2\leq 0\; ,\\\\\frac{t^2+2t+1}{t}\leq 0\; ,\; \; \frac{(t+1)^2}{t}\leq 0\; \; \Rightarrow \; \; t\in \{-1\}\cup (0,+\infty )\; ,\\\\t=-1\; ,\; \; 5x^2-14x-3=-1\; ,\; \; 5x^2-14x-2=0\; ,\\\\D/4=7^2+5\cdot 2=59\; ,\; \; x_{1,2}=\frac{7\pm \sqrt{59}}{5}\\\\x_1\approx -0,136\; \; ,\; \; x_2\approx 2,936$

0\; \; \to \; \; 5x^2-14x-3>0\; ,\; x\in (-\infty ;\, -0,2)\cup (3,+\infty )\\\\Otvet:\; \; x\in (-\infty ;\, -0,2)\cup \{\frac{7-\sqrt{59}}{5}\}\cup\{\frac{7+\sqrt{59}}{5}\}\cup (3,+\infty )\; ." alt="t>0\; \; \to \; \; 5x^2-14x-3>0\; ,\; x\in (-\infty ;\, -0,2)\cup (3,+\infty )\\\\Otvet:\; \; x\in (-\infty ;\, -0,2)\cup \{\frac{7-\sqrt{59}}{5}\}\cup\{\frac{7+\sqrt{59}}{5}\}\cup (3,+\infty )\; ." align="absmiddle" class="latex-formula">