Question

ПОМОГИТЕ ПОЖАЛУЙСТА ДАМ 34 БАЛОВ

Answer 1

$1)\; \; (x+4)(x-6)\geq 0\; ,\qquad +++[-4]---[\, 6\, ]+++\\\\x\in (-\infty ,-4\, ]\cup ]\, 6,+\infty )\\\\2)\; \; -x^2+3,5x+2\geq 0\; |\cdot (-2)\; \; \to \; \; 2x^2-7x-4\leq 0\\\\D=81\; ,\; x_1=-\frac{1}{2}\; \; ,\; \; x-2=4\\\\(x+\frac{1}{2})(x-4)\geq 0\qquad +++[-\frac{1}{2}\, ]---[\;\, 4\, ]+++\\\\x\in (-\infty ,-\frac{1}{2}\, ]\cup [\, 4,+\infty )\\\\Otvet:\; \; -1\; ,\; 2\in (-\infty ,-\frac{1}{2}\, ]\cup [\, 4,+\infty )\; .\\\\3)\; \; -x^2+9x-14>0\; |\cdot (-1)\; \; \to \; \; x^2-9x+14<0\\\\D=25\; ,\; \; x_1=2\; ,\; x_2=7$

$+++(2)---(7)+++\\\\x\in (2,7)\; \; \Rightarrow \; \; celue\; x:\; \; x=3,4,5,6\; .\\\\4)\; \; \left \{ {{x^2-3x-4>0} \atop {3x+12>0}} \right.\; \left \{ {{(x+4)(x-1)>0} \atop {3(x+4)>0}} \right. \; \left \{ {{x\in (-\infty ,-4)\cup (1,+\infty )} \atop {x\in (-4,+\infty )}} \right. \; \; \Rightarrow \; \; x\in (1,+\infty )\\\\5)\; \; \frac{4+4x+x^2}{6-5x+x^2}>0\; \; ,\; \; \frac{(x+2)^2}{-(x-1)(x+6)}>0\; \; ,\; \; \frac{(x+2)^2}{(x-1)(x+6)}<0\\\\+++(-6)---(0)---(1)+++\\\\x\in (-6,0)\cup (0,1)$

$6)\; \; y=\sqrt{x-3}+\sqrt{x^2-7x+6}\\\\OOF:\; \; \left \{ {{x-3\geq 0} \atop {x^2-7x+6\geq 0}} \right. \; \left \{ {{x\geq 3\qquad \quad } \atop {(x-1)(x-6)\geq 0}} \right.\; \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{x\geq 3} \atop {x\in (-\infty ,-6)\cup (-1,+\infty )}} \right. \\\\x\in [\, 3,+\infty )$