Question

Решите неравенство :√4-x

Answer 1

Ответ:

Запишем систему, равносильную заданному неравенству .

$\bf \sqrt{4-x} < x+2\ \ \ \Longleftrightarrow \ \ \ \left\{\begin{array}{l}\bf x+2 > 0\\\bf 4-x\geq 0\\\bf 4-x < (x+2)^2\end{array}\right$        

$\left\{\begin{array}{l}\bf x+2 > 0\\\bf 4-x\geq 0\\\bf 4-x < (x+2)^2\end{array}\right\ \ \left\{\begin{array}{l}\bf x > -2\\\bf x\leq 4\\\bf 4-x < x^2+4x+4\end{array}\right\ \ \left\{\begin{array}{l}\bf x > -2\\\bf x\leq 4\\\bf x^2+5x > 0\end{array}\right\ \ \\\\\\\left\{\begin{array}{l}\bf -2 < x\leq 4\\\bf x\, (x+5) > 0\end{array}\right\ \ \left\{\begin{array}{l}\bf -2 < x\leq 4\\\bf x\in (-\infty ;-5\, )\cup (\ 0\, ;+\infty \, )\end{array}\right\ \bf \ \ \Rightarrow \ \ x\in (\ 0\, ;\ 4\, ]$  

Ответ:  $\bf x\in (\ 0\ ;\ 4\ ]$  .