Question

Допоможіть будь ласка, даю 50 балів​

Answer 1

$\displaystyle\bf\\\sqrt{3x+x^{2} } < 4-x\\\\\\\left\{\begin{array}{ccc}4-x > 0\\3x+x^{2} \geq 0\\3x+x^{2} < (4-x)^{2} \end{array}\right \\\\\\\left\{\begin{array}{ccc}-x > -4\\x\cdot(3+x)\geq 0\\3x+x^{2} < 16-8x+x^{2} \end{array}\right \\\\\\ \left\{\begin{array}{ccc}x < 4\\x\in\Big(-\infty \ ; \ -3\Big]\cup\Big[0 \ ; \ +\infty\Big)\\3x+8x < 16\end{array}\right \\\\\\\left\{\begin{array}{ccc}x < 4\\x\in\Big(-\infty \ ; \ -3\Big]\cup\Big[0 \ ; \ +\infty\Big)\\11x < 16\end{array}\right$

$\displaystyle\bf\\\left\{\begin{array}{ccc}x < 4\\x\in\Big(-\infty \ ; \ -3\Big]\cup\Big[0 \ ; \ +\infty\Big)\\x < 1\dfrac{5}{11} \end{array}\right \\\\\\Otvet \ : \ x\in\Big(-\infty \ ; \ -3\Big]\cup\Big[ 0 \ ; \ 1\frac{5}{11} \Big)$