Question

Помогите, умоляю,ааааааааааааааа

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Answer 1

Ответ:

$\left\{\begin{array}{l}x^2+4x-21<0\\4x+3>x-5\end{array}\right\ \ \left\{\begin{array}{l}(x+7)(x-3)<0\\3x>-8\end{array}\right\ \ \left\{\begin{array}{l}x\in (-7\ ;\ 3\ )\\x\in (-\frac{8}{3}\ ;+\infty \, )\end{array}\right\\\\\\\star \ \ x^2+4x-21=0\ \ ,\ \ D=100\ \ ,\ \ x_1=-7\ ,\ x_2=3\ \ ,\\\\x^2+4x-21=(x+7)(x-3)\ \ \star \\\\\\Otvet:\ \ x\in \Big(-\dfrac{8}{3}\ ;\ 3\ \Big)\ \ \ ili\ \ \ \ x\in \Big(-2\dfrac{2}{3}\ ;\ 3\ \Big)$