Question

(5x+2)^2>=(4-2x)^2
помогите решить неравенство

Answer 1

Ответ:

$(5x+2)^2\geq (4-2x)^2\\\\(5x+2)^2-(4-2x)^2\geq 0\\\\\star \ \ a^2-b^2=(a-b)(a+b)\ \ \star \\\\\Big((5x+2)-(4-2x)\Big)\Big((5x+2)+(4-2x)\Big)\geq 0\\\\(7x-2)(3x+6)\geq 0\\\\3\, (7x-2)(x+2)\geq 0\\\\7x-2=0\ \ ,\ \ x=\dfrac{2}{7}\ \ ;\ \ \ \ x+2=0\ \ ,\ \ x=-2\\\\znaki:\ \ +++[-2\, ]---[\frac{2}{7}\, ]+++\\\\x\in \Big(-\infty \, ;\ -2\ \Big]\cup \Big[\ \dfrac{2}{7}\, ,+\infty \, \Big)$