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Question

Solve the following inequality:
|x+3|+xx+2>1

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Solution

|x+3|+xx+2>1
(i)when(x+3)>0
x>3
x+3+xx+21>0
x+3+xx2x+2>0
(x+1)(x+2)>0
xε(,1)(2,)
butsincex>3
xε(3,)
(ii)when(x+3)<0
x<3
|x+3|=(x+3)
x3+xx+21>0
x3+xx2x+1>0
x+5x+2>0
xε(5,2)
butx<3
xε(5,3)
xε(5,3)(3,)


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