Question

Решить неравенства (уровень егэ!)​

Answer 1

$1)\; \; \sqrt{7-x}<\frac{\sqrt{x^3-6x^2+14x-7} }{\sqrt{x-1}}\; \; ,\; \; \; ODZ:\; \left \{ {{1<x\leq 7\; ,\; x\ne 1\quad } \atop {x^3-6x^2+14x-7\geq 0}} \right.\; \\\\\sqrt{x-1}\ne 0\; \; ,\; tak\; kak\; \; \sqrt{x-1}>0\; ,\; to\; \; \sqrt{7-x}\cdot \sqrt{x-1}<\sqrt{x^3-6x^2+14x-7}\; ,\\\\\sqrt{(7-x)(x-1)}<\sqrt{x^3-6x^2+14x-7}\; \; \Leftrightarrow \; \; \left \{ {{(7-x)(x-1)\geq 0\; ,\; x\ne 1 \quad } \atop {(7-x)(x-1)<x^3-6x^2+14x-7}} \right.\\\\a)\; \; (7-x)(x-1)<x^3-6x^2+14x-7\\\\-x^2+8x-7<x^3-6x^2+14x-7$

$x^3-5x^2+6x>0\; \; ,\; \; x(x-2)(x-3)>0\; ,\\\\znaki:\; ---(0)+++(2)---(3)+++\\\\x\in (0,2)\cup (3,+\infty )\\\\b)\; \; \left \{ {{1<x\leq 7\qquad \quad } \atop {x\in (0,2)\cup (3,+\infty )}} \right.\; \; \Rightarrow \; \; x\in (1,2)\cup (3,7\, ]$

$2)\; \; 1)\; \; \sqrt{5-x}<\frac{\sqrt{x^3-7x^2+14x-5} }{\sqrt{x-1}}\; \; ,\; \; \; ODZ:\; \left \{ {{1<x\leq 5\qquad \qquad } \atop {x^3-7x^2+14x-7\geq 0}} \right.\; \\\\\sqrt{x-1}\ne 0\; \; ,\; tak\; kak\; \; \sqrt{x-1}>0\; ,\; to\; \; \sqrt{5-x}\cdot \sqrt{x-1}<\sqrt{x^3-7x^2+14x-5}\; ,\\\\\sqrt{(5-x)(x-1)}<\sqrt{x^3-7x^2+14x-5}\; \; \Leftrightarrow \; \; \left \{ {{(5-x)(x-1)\geq 0\; ,\; x\ne 1 \qquad \; \; \; \; } \atop {(5-x)(x-1)<x^3-7x^2+14x-5}} \right.\\\\a)\; \; (5-x)(x-1)<x^3-7x^2+14x-5$

$-x^2+6x-5<x^3-7x^2+14x-5\\\\x^3-6x^2+8x>0\; \; ,\; \; x(x^2-6x+8)>0\; \; ,\; \; x(x-2)(x-4)>0\; ,\\\\znaki:\; \; ---(0)+++(2)---(4)+++\\\\x\in (0,2)\cup (4,+\infty )\\\\b)\; \; \left \{ {{1<x\leq 5\qquad \quad } \atop {x\in (0,2)\cup (4,+\infty )}} \right.\; \; \Rightarrow \; \; \; x\in (1,2)\cup (4,5\; ]$