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Since `x^2>=0` and the fraction `x^2/(x+2)` is positive, it must be the case that `x+2>0,` or equivalently `x> -2.` This means we can multiply both sides by `x+2` without having to worry about the inequality sign flipping. We get
`x^2>x+2,` or `x^2-x-2>0.` Factor to get
`(x+1)(x-2)>0.` This product will be positive whenever the factors are either both positive or both negative. Solving `x+1>0` and `x-2>0` simultaneously gives `x> -1` and `x> 2.` The first one is redundant, so both factors are positive and the inequation is true whenever `x> 2`.
Now solving `x+1<0` and `x-2<0` simultaneously gives `x<-1` and `x<2.` Here, if they're both true, the second one is redundant, and both factors are negative and the inequation is true whenever `-2<x< -1` (remember at the very beginning we said that `-2<x` ).
Combining the two solutions gives `(-2,-1)uu(2,oo)` as the solution set.
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