(13)|x+2|2−|x|>9
⇒3|x+2|2−|x|>32
⇒|x+2|2−|x|>2
⇒|x+2|2−|x|<−2
Case I: x<−2
−(x+2)2+x<−2
⇒−1<−2, which is not true.
Case II: −2<x<0
(x+2)2+x<−2
⇒1<−2, which is not true.
Case III: x≥0
x+22−x<−2
⇒x+22−x+2<0
⇒6−x2−x<0
⇒2<x<6
Hence, the solution is (2,6).