Question

Решить неравенство:
X^3 -1/x^3 >= 4*(x-1/x)

Answer 1

я тут все пошагово сфотографировала

Answer 2

Ответ:

$x^3-\dfrac{1}{x^3}\geq 4\Big(x-\dfrac{1}{x}\Big)\ \ ,\ \ \ ODZ:\ x\ne 0\ ,\\\\\\t=x-\dfrac{1}{x}\ \ ,\ \ \ t^3=x^3-3x^2\cdot \dfrac{1}{x}+3x\cdot \dfrac{1}{x^2}-\dfrac{1}{x^3}=x^3-\dfrac{1}{x^3}-3x+\dfrac{3}{x}=\\\\\\{}\qquad \qquad \qquad \qquad =\Big(x^3-\dfrac{1}{x^3}\Big)-3\Big(x-\dfrac{1}{x}\Big)=\Big(x^3-\dfrac{1}{x^3}\Big)-3t\ \ ,\\\\x^3-\dfrac{1}{x^3}=t^3+3t\ \ ,\\\\t^3+3t\geq 4t\ \ ,\ \ \ t^3-t\geq 0\ \ ,\ \ \ t\, (t^2-1)\geq 0\ \ ,\ \ t\, (t-1)(t+1)\geq 0\ \ ,\\\\znaki:\ \ ---[-1]+++[0]---[1]+++$  

$t\in [\ -1\ ;\ 0\ ]\cup [\ 1\ ;+\infty )\\\\a)\ \ -1\leq t\leq 0\ \ ,\ \ -1\leq x-\dfrac{1}{x}\leq 0\ \ ,\ -1\leq \dfrac{x^2-1}{x}\leq 0\ \ ,\\\\\dfrac{x^2-1}{x}\leq 0\ \ ,\ \ \dfrac{(x-1)(x+1)}{x}\leq 0\ \ ,\ \ znaki:\ \ ---[-1]+++(0)---[\ 1\ ]+++\\\\x\in (-\infty ;-1\ ]\cup (\ 0\ ;1\ ]\\\\\dfrac{x^2-1}{x}\geq -1\ \ ,\ \ \dfrac{x^2-1+x}{x}\geq 0\ \ ,\ \ \dfrac{(x-x_1)(x-x_2)}{x}\geq 0\\\\x^2+x-1=0\ \ ,\ \ D=5\ \ ,\ \ x_1=\dfrac{-1-\sqrt5}{2}\approx -1,62\ ,\ \ x_2=\dfrac{-1+\sqrt5}{2}\approx 0,62$

$znaki:\ \ ---[x_1]+++(0)---[x_2]+++\\\\x\in \Big[\ \dfrac{-1-\sqrt5}{2} \ ;\ 0\ \Big)\cup \Big[\dfrac{-1+\sqrt5}{2}\ ;+\infty \Big)\\\\\\\left\{\begin{array}{l}x\in (-\infty ;-1\ ]\cup (\ 0\ ;\ 1\ ]\\x\in \Big[\ \dfrac{-1-\sqrt5}{2} \ ;\ 0\ \Big)\cup \Big[\dfrac{-1+\sqrt5}{2}\ ;+\infty \Big)\end{array}\right\ \ \ \Rightarrow \\\\\\x\in \Big[\dfrac{-1-\sqrt5}{2}\ ;-1\ \Big]\cup \Big[ \dfrac{-1+\sqrt5}{2}\ ;\ 1\ \Big]$

$b)\ \ t\geq 1\ \ ,\ \ x-\dfrac{1}{x}\geq 1\ \ ,\ \ \dfrac{x^2-x-1}{x}\geq 0\ \ ,\dfrac{(x-x_3)(x-x_4)}{x}\geq 0\ ,\\\\\\x^2-x-1=0\ \ ,\ \ x_{3}=\dfrac{1-\sqrt5}{2}\approx -0,62\ ,\ \ x_2=\dfrac{1+\sqrt5}{2}\approx 1,62\\\\znaki:\ \ ---[x_3]+++(0)---[x_4]+++\\\\x\in \Big[\ \dfrac{1-\sqrt5}{2}\ ;\ 0\ \Big)\cup \Big[\ \dfrac{1+\sqrt5}{2}\ ;+\infty \Big)$

$Otvet:\\\\x\in \Big[\ \dfrac{-1-\sqrt5}{2}\ ;-1\ \Big]\cup \Big[\ \dfrac{1-\sqrt5}{2}\ ;\ 0\ \Big)\cup \Big[\ \dfrac{-1+\sqrt5}{2}\ ;\ 1\ \Big]\cup \Big[\ \dfrac{1+\sqrt5}{2}\ ;+\infty \Big)$