Question

ДАЮ 30 БАЛЛОВ!!!! ПОЖАЛУЙСТА, АЛГЕБРА, ЛОГАРИФМЫ

РЕШИТЕ, пожалуйста

Answer 1

Ответ:

$log_{15}(x+5)+log_{15}(x^2+\dfrac{1}{x+5}\, )\leq 2\cdot log_{15}\dfrac{x^2+x+5}{2}\ \ ,\ \ ODZ:\ x>-5\ ,\\\\\\log_{15}\Big((x+5)\cdot (x^2+\dfrac{1}{x+5}\, )\Big)\leq log_{15}\Big(\dfrac{x^2+x+5}{2}\Big)^2\\\\log_{15}\Big((x+5)\cdot \dfrac{x^2(x+5)+1}{x+5}\, \Big)\leq log_{15}\dfrac{(x^2+(x+5)\, )^2}{4}\\\\\\log_{15}(x^2(x+5)+1)\leq log_{15}\dfrac{(x^2)^2+2x^2(x+5)+(x+5)^2}{4}\\\\\\a=15>1\ \ \Rightarrow \ \ x^2(x+5)+1\leq \dfrac{(x^2)^2+2x^2(x+5)+(x+5)^2}{4}$

$4x^2(x+5)+4\leq (x^2)^2+2x^2(x+5)+(x+5)^2\\\\\underbrace{(x^2)^2-2x^2(x+5)+(x+5)^2}-4\geq 0\\\\\Big(x^2-(x+5)\Big)^2-2^2\geq 0\\\\(x^2-(x+5)-2)(x^2-(x+5)+2)\geq 0\\\\(x^2-x-7)(x^2-x-3)\geq 0\\\\\star \ \ x^2-x-7=0\ \ ,\ \ x_{1}=\dfrac{1-\sqrt{29}}{2}\approx -2,2\ \ ,\ \ x_2=\dfrac{1+\sqrt{29}}{2}\approx 3,2\ \ \star \\\\\star \ \ x^2-x-3=0\ \ ,\ \ x_{3}=\dfrac{1-\sqrt{13}}{2}\approx -1,3\ \ ,\ \ x_4=\dfrac{1+\sqrt{13}}{2}\approx 2,3\ \ \star$

$\Big(x-\dfrac{1-\sqrt{29}}{2}\Big)\Big(x-\dfrac{1+\sqrt{29}}{2}\Big)\Big(x-\dfrac{1-\sqrt{13}}{2}\Big)\Big(x-\dfrac{1+\sqrt{13}}{2}\Big)\geq 0\\\\\\\star \ \ (x+2,2)(x-3,19)(x+1,2)(x-2,3)\geq 0\\\\znaki:\ \ (-5)+++[-2,2]---[-1,3]+++[2,3]---[3,2]+++\\\\x\in (-5\ ;-2,2\ ]\cup [\, -1,3\ ;\ 2,3\ ]\cup [\ 3,19\ ;+\infty )$

$Otvet:\ x\in \Big(-5\ ;\dfrac{1-\sqrt{29}}{2}\, \Big]\cup \Big[\, \dfrac{1-\sqrt{13}}{2}\ ;\ \dfrac{1+\sqrt{13}}{2}\, \Big]\cup \Big[\, \dfrac{1+\sqrt{29}}{2}\ ;+\infty \Big)\ .$