Question

решить все неравенства

Answer 1

Ответ:

$1)\ \ \dfrac{x^2}{5}\geq \dfrac{1,4x-0,6}{2}\ \Big|\cdot 10\\\\2x^2\geq 5(1,4x-0,6)\ \,\ \ \ 2x^2-7x+3\geq 0\ \ ,\ \ D=25\ ,\ x_1=0,5\ ,\ x_2=3\\\\2(x-0,5)(x-3)\geq 0\\\\znaki:\ \ \ +++[\ 0,5\ ]---[\ 3\ ]+++\\\\x\in (-\infty ;\ 0,5\ ]\cup [\ 3\, ;+\infty \, )\\\\\\2)\ \ (\underbrace {\sqrt5-2,1}_{>0})(5+2x)>0\ \ ,\ \ \ 5+2x>0\ \ ,\ \ 2x>-5\ \ ,\ \ x>-2,5\ \ ,\\\\x\in (-2,5\ ;\ +\infty )$

$3)\ \ x^2(-x^2-81)\leq 81(-x^2-81)\\\\x^2(-x^2-81)-81(-x^2-81)\leq 0\\\\(-x^2-81)(x^2-81)\leq 0\ \ ,\ \ -(x^2+81)(x^2-81)\leq 0\ \ ,\\\\(x^2+81)(x^2-81)\geq 0\ \ ,\ \ (x^2+81)(x-9)(x+9)\geq 0\ \ ,\\\\znaki:\ \ +++[-9\ ]---[\ 9\ ]+++\\\\x\in (-\infty \, ;-9\ ]\cup [\ 9\, ;+\infty \, )$

$4)\ \ (x-3)^2>\sqrt5(x-3)\\\\(x-3)^2-\sqrt5(x-3)>0\ \ ,\ \ \ \ (x-3)(x-3-\sqrt5)>0\ \ ,\\\\znaki:\ \ +++(3)---(3+\sqrt5)+++\\\\x\in (-\infty \, ;\, 3\, )\cup (\, 3+\sqrt5\ ;+\infty \, )$

$5)\ \ (2x-3)^2\geq (3x-2)^2\\\\(2x-3)^2-(3x-2)^2\geq 0\\\\\Big((2x-3)-(3x-2)\Big)\Big((2x-3)+(3x-2)\Big)\geq 0\\\\(-x-1)(5x-5)\geq 0\ \Big|\cdot (-1)\\\\(x+1)(5x-5)\leq 0\\\\x+1=0\ \ \to \ \ x=-1\ \ ;\ \ \ \ 5x-5=0\ \ \to \ \ \ x=1\ ;\\\\znaki:\ \ +++[-1\ ]---[\ 1\ ]+++\\\\x\in [-1\ ;\ 1\ ]$