Question

решите квадратное неравинство
даю 40 балов
МОМОГИТЕ ПОЖАЛУЙСТА
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Answer 1

Ответ:

$10)\ \ 7x^2+18x<-5\\\\7x^2+18x+5=0\ \ ,\ \ D/4=46\ \ ,\\\\x_1=\dfrac{-9-\sqrt{46}}{7}\approx -2,25\ \ \ ,\ \ \ x_2=\dfrac{-9+\sqrt{46}}{7}\approx -0,32\\\\znaki:\ \ \ \ +++(x_1)---(x_2)+++\\\\x\in \Big(\ \dfrac{-9-\sqrt{46}}{7}\ ;\ \dfrac{-9+\sqrt{46}}{7}\ \Big)$

$12)\ \ 9x^2-4x-2\geq 0\\\\D/4=22\ \ ,\ \ \ x_1=\dfrac{2-\sqrt{22}}{\frac{x}{y} 9}\ \ ,\ \ \ x_2=\dfrac{2+\sqrt{22}}{9}\\\\znaki:\ \ \ \ +++(x_1)---(x_2)+++\\\\x\in \Big(-\infty \, ;\ \dfrac{2-\sqrt{22}}{9}\ \Big]\cup \Big[\ \dfrac{2+\sqrt{22}}{9}\ ;\ +\infty \, \Big)$

$14)\ \ 3y^2-6y+3>0\ \ ,\ \ \ y^2-2y+1>0\ \ ,\ \ \ (y-1)^2>0\ \ \Rightarrow \ \ (y-1)\ne 0\ ,\\\\y\ne 1\ \ \Rightarrow \ \ \ y\in (-\infty \ ;\ 1\ )\cup (\ 1\ ;+\infty \, )\\\\\\15)\ \ 2y^2+9y-486\leq 0\\\\D=3969=63^2\ \ ,\ \ \ y_1=\dfrac{-9-63}{4}=-18\ \ ,\ \ x_2=\dfrac{-9+63}{4}=13,5\\\\znaki:\ \ \ +++(-18)---(13,5)+++\\\\y\in [-18\ ;\ 13,5\ ]$