Question

помогите пожалуйста​

Answer 1

Ответ:

$1)\ \ \left\{\begin{array}{l}4x^2-5x-7>0\\x\geq 3\end{array}\right\ \ \left\{\begin{array}{l}x\in \Big(-\infty ;\ \dfrac{5-\sqrt{137}}{8}\ \Big)\cup \Big(\ \dfrac{5+\sqrt{137}}{8}\ ;+\infty \Big)\\x\in [\ 3;+\infty \, )\end{array}\right\ \ \\\\\\4x^2-5x-7=0\ \ ,\ \ D=137\ ,\\\\x_1=\dfrac{5-\sqrt{137}}{8}\approx -0,84\ ,\ x_2=\dfrac{5+\sqrt{137}}{8}\approx 2,09$

$4x^2-5x-7>0\ \ \to \ \ x\in \Big(-\infty ;\ \dfrac{5-\sqrt{137}}{8}\ \Big)\cup \Big(\ \dfrac{5+\sqrt{137}}{8}\ ;+\infty \Big)\\\\\\Otvet:\ x\in [\ 3\ ;+\infty \, )\ .$

$2)\ \ \left\{\begin{array}{l}x^2-10x+16>0\\x\leq 3\end{array}\right\ \ \left\{\begin{array}{l}(x-2)(x-8)>0\\x\leq 3\end{array}\right\ \ \left\{\begin{array}{l}x\in (-\infty ;2\ )\cup (\ 8\ ;+\infty \, )\\x\in (-\infty ;\ 3\ ]\end{array}\right\\\\\\x^2-10x+16=0\ \ ,\ \ D/4=9\ \ ,\ \ x_1=2\ ,\ \ x_2=8\\\\Otvet:\ x\in (-\infty ;\ 2\ )$

$3)\ \ \left\{\begin{array}{l}x^2-23x-24\leq 0\\2x+5\geq x-3\end{array}\right\ \ \left\{\begin{array}{l}(x+1)(x-24)\leq 0\\x\geq -8\end{array}\right\ \ \left\{\begin{array}{l}x\in [-1\ ;\ 24\ ]\\x\in [-8\ ;+\infty \, )\end{array}\right\\\\\\x^2-23x-24=0\ \ ,\ \ D=625\ ,\ \ x_1=-1\ ,\ x_2=24\\\\Otvet:\ \ x\in [-1\ ;\ 24\ ]\ .$

$4)\ \ \left\{\begin{array}{l}x^2-14x-32<0\\4x+1\leq 2x+5\end{array}\right\ \ \left\{\begin{array}{l}(x+2)(x-16)<0\\2x\leq 4\end{array}\right\ \ \left\{\begin{array}{l}x\in (-2\ ;\ 16\ )\\x\in (-\infty ;\ 2\ ]\end{array}\right\\\\\\x^2-14x-32=0\ \ ,\ \ D/4=81\ \ ,\ \ x_1=-2\ ,\ x_2=16\\\\Otvet:\ \ x\in (-2\ ;\ 2\ ]\ .$