Question

Решите неравенство
50 баллов, срочно !!!

Answer 1

$\frac{\sqrt{x^2+x-6}+3x+13}{x+5}\ \textgreater \ 1\; \; \; \to \; \; \; \frac{\sqrt{x^2+x-6}+3x+13}{x+5}-1\ \textgreater \ 0\\\\\frac{\sqrt{x^2+x-6}+3x+13-x-5}{x+5}\ \textgreater \ 0\; \; ,\; \; \; \frac{\sqrt{x^2+x-6}+2x+8}{x+5} \ \textgreater \ 0\; \; \Rightarrow \\\\a)\; \; \left \{ {{\sqrt{x^2+x-6}+2x+8\ \textgreater \ 0} \atop {x+5\ \textgreater \ 0}} \right. \; \; \; ili\; \; \; b)\; \left \{ {{\sqrt{x^2+x-6}+2x+8\ \textless \ 0} \atop {x+5\ \textless \ 0}} \right. \\\\a)\; \; \sqrt{x^2+x-6}+2x+8\ \textgreater \ 0\; \; ,\; \; \sqrt{x^2+x-6}\ \textgreater \ -2x-8\; \; \Leftrightarrow$

$\left \{ {{-2x-8 \geq 0} \atop {x^2+x-6\ \textgreater \ 4x^2+32x+64}} \right. \; \; \; ili\; \; \; \left \{ {{-2x-8\ \textless \ 0} \atop {x^2+x-6 \geq 0}} \right. \\\\ \left \{ {{-2x \geq 8} \atop {3x^2+31x+70\ \textless \ 0}} \right. \; \; \; ili\; \; \; \left \{ {{-2x\ \textless \ 8} \atop {(x-2)(x+3) \geq 0}} \right. \\\\\left \{ {{x \leq 4} \atop {(x+7)(x+\frac{10}{3})\ \textless \ 0}} \right. \; \; \; ili\; \; \; \left \{ {{x\ \textgreater \ -4} \atop {x\in (-\infty ,-3\, ]\cup [\, 2,+\infty )}} \right. \\\\\left \{ {{x \leq 4} \atop {x\in (-7,-\frac{10}{3})}} \right. \; \; \; ili\; \; \; x\in (-4,-3\, ]$

$x\in (-7,-\frac{10}{3})\; \; ili\; \; x\in (-4,-3\, ]\; \; \Rightarrow \\\\x\in (-7,-\frac{10}{3})\cup ({-\frac{10}{3},-3\, ]$

$x+5\ \textgreater \ 0\; \; \Rightarrow \; \; x\ \textgreater \ -5\\\\ \left \{ {{x\in (-7,-\frac{10}{3})\cup (-\frac{10}{3},-3\, ]} \atop {x\in (-5,+\infty )}} \right. \; \; \Rightarrow \ \; \; \underline {x\in (-5,-\frac{10}{3})\cup (-\frac{10}{3},-3\, ]}\\\\b)\; \; \sqrt{x^2+x-6}\ \textless \ -2x-8\; \; \Leftrightarrow \; \; \; \left \{ {{-2x-8\ \textgreater \ 0\; ,\; x^2+x-6 \geq 0} \atop {x^2+x-6\ \textless \ 4x^2+32x+64}} \right. \\\\ \left \{ {{x\ \textless \ -4\; ,\; x\ \textgreater \ 2,\; x\ \textless \ -3} \atop {3x^2+31x+70\ \textgreater \ 0}} \right. \; \left \{ {{x\in (-\infty ,-4)} \atop {x\in (-\infty ,-7)\cup (-\frac{10}{3},+\infty )}} \right. \; \; \Rightarrow$

$x\in (-\infty ,-7)\\\\x+5<0\; \; \Rightarrow \; \; x<-5\\\\\left \{ {{x\ \textless \ -5} \atop {x\in (-\infty ,-7)}} \right. \; \; \Rightarrow \; \; \underline {x\in (-\infty ,-7)}\\\\Otvet:\; \; x\in (-\infty ,-7)\cup (-5,-\frac{10}{3})\cup(-\frac{10}{3},-3\, ]\; .$