Подробное решение неравенств

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

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

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

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

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

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

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

<img src="https://tex.z-dn.net/?f=4%29%5Cquad++%5Cfrac%7Bx-6%7D%7B%5Csqrt%7Bx%2B9%7D%7D+%5C+%5Ctextgreater+%5C+2%5C%3B+%2C%5C%3B+%5C%3B+%5C%3B+ODZ%3A%5C%3B+x%5C+%5Ctextgreater+%5C+-9%5C%5C%5C%5C+%5Cfrac%7Bx-6-2%5Csqrt%7Bx%2B9%7D%7D%7B%5Csqrt%7Bx%2B9%7D%7D+%5C+%5Ctextgreater+%5C+0%5C%3B+%5C%3B+%5Cto+%5C%3B+%5C%3B+x-6-2%5Csqrt%7Bx%2B9%7D%5C+%5Ctextgreater+%5C+0%5C%5C%5C%5C2%5Csqrt%7Bx%2B9%7D%5C+%5Ctextless+%5C+x-6%5C%3B+%5C%3B+%5CLeftrightarrow+%5C%3B+%5C%3B++%5Cleft+%5C%7B+%7B%7Bx%2B9%5C+%5Ctextgreater+%5C+0%5C%3B+%2C%5C%3B+x-6%5C+%5Ctextgreater+%5C+0%7D+%5Catop+%7B4%28x%2B9%29%5C+%5Ctextless+%5C+%28x-6%29%5E2%7D%7D+%5Cright.+%5C%5C%5C%5C+%5Cleft+%5C%7B+%7B%7Bx%5C+%5Ctextgreater+%5C+-9%5C%3B+%2C%5C%3B+x%5C+%5Ctextgreater+%5C+6%7D+%5Catop+%7B4x%2B36%5C+%5Ctextless+%5C+x%5E2-12x%2B36%7D%7D+%5Cright.+%5C%3B++%5Cleft+%5C%7B+%7B%7Bx%5C+%5Ctextgreater+%5C+6%7D+%5Catop+%7Bx%5E2-16x%5C+%5Ctextgreater+%5C+0%7D%7D+%5Cright.+%5C%3B++%5Cleft+%5C%7B+%7B%7Bx%5C+%5Ctextgreater+%5C+6%7D+%5Catop+%7Bx%28x-16%29%5C+%5Ctextgreater+%5C+0%7D%7D+%5Cright.+%5C%3B++%5Cleft+%5C%7B+%7B%7Bx%5C+%5Ctextgreater+%5C+6%7D+%5Catop+%7Bx%5C+%5Ctextless+%5C+0%5C%3B+ili%5C%3B+x%5C+%5Ctextgreater+%5C+16%7D%7D+%5Cright.+%5C%3B+%5CRightarrow+%5C%5C%5C%5Cx%5Cin+%2816%2C%2B%5Cinfty+%29%5C%3B+-%5C%3B+otvet" id="TexFormula8" title="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}} \

NNNLLL54_zn (832k баллов) 09 Май, 18