Question

[tex] \frac{ x^{2} }{x+2} \geq 1
Ajutor!!! Va rog [/tex]

Answer 1

[tex]\dfrac{x^2}{x+2}\geq 1\\ \dfrac{x^2}{x+2}-1\geq 0\\ \dfrac{x^2-x-2}{x+2}\geq 0\\ \text{Mai departe se face cu tabel de semn:}\\ x^2-x-2=0\\ \Delta=1+8=9\Rightarrow \sqrt{\Delta}=3\\ x_1=\dfrac{1+3}{2}=2\\ x_2=\dfrac{1-3}{2}=-1\\ x+2=0\Rightarrow x=-2\\ ~~~~~~x~~~~~~~ |-\infty ~~~~~~~~~ -2~~~~~~~-1~~~~~~~~~2~~~~~~~~\infty \\ ~~~~x+2~~~~| -~~~-~~-~~~ 0 +~~~~~~~+~~~~~+~~~~~+ \\ x^2-x-2| +~~~~+~~~~+~~~~~+~~~0-~~-~0+~~~~+~~~~+~~~~~+\\ ~~~~ sgn~~~~~~| -~~~~-~~~-~ 0+~~~+/-~~-~/+~~~+~~~~+~~~~+\\ [/tex]
$S:x\in [-2;-1)\cup (2,\infty)$