Подробное решение неравенств

Question

Алгебра

Светлана

7 год назад

Подробное решение неравенств

ОТВЕТЫ

Ксенофонтович

Jul 5, 2019

$1)\quad \frac{x+3}{\sqrt{x+1}}\ \textgreater \ 3\; ,\; \; ODZ:\; \; x+1\ \textgreater \ 0\; ,\; \; x\ \textgreater \ -1\\\\\frac{x+3-3\sqrt{x+1}}{\sqrt{x+1}}\ \textgreater \ 0\\\\Tak\; kak\; \sqrt{x+1}\ \textgreater \ 0,\; to\; \; \; \; x+3-3\sqrt{x+1}\ \textgreater \ 0\; ,\\\\3\sqrt{x+1}\ \textless \ x+3\; \; \; \Rightarrow \; \; \; \left \{ {{x+1\ \textgreater \ 0\; ,\; x+3\ \textgreater \ 0} \atop {9(x+1)\ \textless \ (x+3)^2}} \right. \\\\ \left \{ {{x\ \textgreater \ -1} \atop {9x+9\ \textless \ x^2+6x+9}} \right. \; \left \{ {{x\ \textgreater \ -1} \atop {x^2-3x\ \textgreater \ 0}} \right. \; \left \{ {{x\ \textgreater \ -1} \atop {x(x-3)\ \textgreater \ 0}} \right. \; \left \{ {{x\ \textgreater \ -1} \atop {x\ \textless \ 0\; ili\; x\ \textgreater \ 3}} \right.$

$x\in (-1,0)\cup (3,+\infty )$

$2)\quad \frac{x+4}{\sqrt{x+1}}\ \textless \ 4\; ,\; \; ODZ:\; \; x\ \textgreater \ -1\\\\\frac{x+4-4\sqrt{x+1}}{\sqrt{x+1}}\ \textless \ 0\\\\\sqrt{x+1}\ \textgreater \ 0\; \; \to \; \; \; x+4-4\sqrt{x+1}\ \textless \ 0\\\\4\sqrt{x+1}\ \textgreater \ x+4\; \; \; \Leftrightarrow \; \; \left \{ {{x+4\ \textless \ 0} \atop {x+1\ \textgreater \ 0}} \right. \; ili\; \; \left \{ {{x+4 \geq 0} \atop {16(x+1)\ \textgreater \ (x+4)^2}} \right. \\\\ \left \{ {{x\ \textless \ -4} \atop {x\ \textgreater \ -1}} \right. \; \; ili\; \; \left \{ {{x \geq -4} \atop {16x+16\ \textgreater \ x^2+8x+16}} \right.$

$x\in \varnothing \; \; \; ili\; \; \; \left \{ {{x \geq -4} \atop {x^2-8xlt;0}} \right.$

$\qquad \quad \left \{ {{x \geq -4} \atop {x(x-8)\ \textless \ 0}} \right. \; \left \{ {{x \geq -4} \atop {0\ \textless \ x\ \textless \ 8}} \right. \; \; \to \; \; x\in (0,8)\; \; -\; otvet$

$3)\quad \frac{x-9}{\sqrt{x+9}} \ \textless \ 1\; ,\; \; \; ODZ:\; x\ \textgreater \ -9 \\\\ \frac{x-9-\sqrt{x+9}}{\sqrt{x+9}} \ \textless \ 0\; \; \to \; \; x-9-\sqrt{x+9}\ \textless \ 0\\\\\sqrt{x+9}\ \textgreater \ x-9\; \; \Leftrightarrow \; \; \left \{ {{x-9\ \textless \ 0} \atop {x+9\ \textgreater \ 0}} \right. \; \; ili\; \; \left \{ {{x-9 \geq 0} \atop {x+9\ \textgreater \ (x-9)^2}} \right. \\\\ \left \{ {{x\ \textless \ 9} \atop {x\ \textgreater \ -9}} \right. \; \; ili\; \; \left \{ {{x \geq 9} \atop {x+9\ \textgreater \ x^2-18x+81}} \right. \\\\-9\ \textless \ x\ \textless \ 9\; \; ili\; \; \left \{ {{x \geq 9} \atop {x^2-19x+72\ \textless \ 0}} \right. \; \to$

$\left \{ {{x \geq 9} \atop {\frac{1}{2}(19-\sqrt{73})\ \textless \ x\ \textless \ \frac{1}{2}(19+\sqrt{73})}} \right. \; \; \to \; \; x\in [\, 9,\; \frac{1}{2}(19+\sqrt{73})\; )\\\\Otvet:\; \; x\in (-9,\; \frac{1}{2}(19+\sqrt{73})\; )$

$4)\quad \frac{x-6}{\sqrt{x+9}} \ \textgreater \ 2\; ,\; \; \; ODZ:\; x\ \textgreater \ -9\\\\ \frac{x-6-2\sqrt{x+9}}{\sqrt{x+9}} \ \textgreater \ 0\; \; \to \; \; x-6-2\sqrt{x+9}\ \textgreater \ 0\\\\2\sqrt{x+9}\ \textless \ x-6\; \; \Leftrightarrow \; \; \left \{ {{x+9\ \textgreater \ 0\; ,\; x-6\ \textgreater \ 0} \atop {4(x+9)\ \textless \ (x-6)^2}} \right. \\\\ \left \{ {{x\ \textgreater \ -9\; ,\; x\ \textgreater \ 6} \atop {4x+36\ \textless \ x^2-12x+36}} \right. \; \left \{ {{x\ \textgreater \ 6} \atop {x^2-16x\ \textgreater \ 0}} \right. \; \left \{ {{x\ \textgreater \ 6} \atop {x(x-16)\ \textgreater \ 0}} \right. \; \left \{ {{x\ \textgreater \ 6} \atop {x\ \textless \ 0\; ili\; x\ \textgreater \ 16}} \right. \; \Rightarrow \\\\x\in (16,+\infty )\; -\; otvet$

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