Question

|x^2+x-2|+|x+4|<=x^2+2x+6

Answer 1

раскрываем модули и получаем систему:

$\\\left[\begin{array}{cccc}\left \{\begin{array}{ccc}x^2+x-2\geq 0\\x+4\geq 0\\x^2+x-2+x+4\leq x^2+2x+6\end{array}\right.\\\left \{\begin{array}{ccc}x^2+x-2<0\\x+4\geq 0\\-x^2-x+2+x+4\leq x^2+2x+6\end{array}\right.\\\left \{\begin{array}{ccc}x^2+x-2< 0\\x+4< 0\\-x^2-x+2-x-4\leq x^2+2x+6\end{array}\right.\\ \left \{\begin{array}{ccc}x^2+x-2\geq 0\\x+4<0\\x^2+x-2-x-4\leq x^2+2x+6\end{array}\right. \end{array}\right.$

$\\\left[\begin{array}{cccc}\left \{\begin{array}{ccc}x^2+x-2\geq 0\\x+4\geq 0\\2\leq 6\end{array}\right.\\\left \{\begin{array}{ccc}x^2+x-2<0\\x+4\geq 0\\x^2+x\geq 0\end{array}\right.\\\left \{\begin{array}{ccc}x^2+x-2< 0\\x+4< 0\\x^2+2x+4\geq 0\end{array}\right.\\ \left \{\begin{array}{ccc}x^2+x-2\geq 0\\x+4<0\\x+6\geq 0\end{array}\right. \end{array}\right.$

решаем:

$1)\left \{\begin{array}{ccc}x^2+x-2\geq 0\\x+4\geq 0\\2\leq 6\end{array}\right.\\x^2+x-2\geq 0\\D=1+8=9=3^2\\x_1=\frac{-1+3}{2}=1\\x_2=\frac{-1-3}{2}=-2$

 +         -          +

----[-2]------[1]------>x

$x \in (-\infty;-2]\cup [1;+\infty)$

$x+4\geq 0\\x\geq -4\\x\in [-4;+\infty)\\\\2\leq 6\\x \in R\\\left \{\begin{array}{ccc}x \in (-\infty;-2]\cup [1;+\infty)\\x\in [-4;+\infty)\\x \in R\end{array}\right. \Rightarrow x\in[-4;-2]\cup [1;+\infty)$

$2)\left \{\begin{array}{ccc}x^2+x-2<0\\x+4\geq 0\\x^2+x\geq 0\end{array}\right.\\x^2+x-2<0\\x\in (-2;1)\\x\geq -4\\x\in [-4;+\infty)\\x(x+1)\geq 0\\x_1=0\\x_2=-1$

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----[-1]-----[0]----->x

$x\in (-\infty;-1]\cup [0;+\infty)$

$\left \{\begin{array}{ccc}x\in (-2;1)\\x\in [-4;+\infty)\\x\in (-\infty;-1]\cup [0;+\infty)\end{array}\right. \Rightarrow x \in(-2;-1]\cup [0;1)$

$3)\left \{\begin{array}{ccc}x^2+x-2< 0\\x+4< 0\\x^2+2x+4\geq 0\end{array}\right.\\x^2+x-2<0\\x\in (-2;1)\\x+4<0\\x\in (-\infty;-4)\\x^2+2x+4\geq 0\\\left \{ {{D=4-16<0} \atop {a>0}} \right. \Rightarrow x^2+2x+4>0,\ \forall x \in R\\x \in R\\\left \{\begin{array}{ccc}x\in (-2;1)\\x\in (-\infty;-4)\\x\in R\end{array}\right. \Rightarrow x \in \varnothing$

$4)\left \{\begin{array}{ccc}x^2+x-2\geq 0\\x+4<0\\x+6\geq 0\end{array}\right.\\x^2+x-2\geq 0\\x \in (-\infty;-2]\cup [1;+\infty)\\x+4<0\\x\in (-\infty;-4)\\x+6\geq 0\\x \in[-6;\+\infty)\\\left \{\begin{array}{ccc}x \in (-\infty;-2]\cup [1;+\infty)\\x\in (-\infty;-4)\\x \in[-6;+\infty)\end{array}\right. \Rightarrow x \in[-6;-4)$

В итоге:

$\left[\begin{array}{cccc}x\in[-4;-2]\cup [1;+\infty)\\x \in(-2;-1]\cup [0;1)\\x \in \varnothing\\ x \in[-6;-4) \end{array}\right. \Rightarrow x \in[-6;-1]\cup [0;+\infty)$

Ответ: $x \in[-6;-1]\cup [0;+\infty)$