2.8 The solution set of - contains (A) (* 1 x 20) (B) (x | x > 0) {-1) (C) (-1, 1) (D) x 1x2 1 or XS-1)

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( frac{x+1}{x}+(x+1)=frac{x+1,2}{1 x} ) we can write ( (x+1)^{2}=|x+1|^{2} ) ( frac{|x+1|}{|x|}+|x+1|=frac{|x+1|^{2}}{|x|} ) ( frac{1}{4} ) ( frac{1}{1 x !}+1 quad=frac{1 times 1+11}{121} ) ( x=- ) ( frac{1+|x|}{|x|}=frac{|x+1|}{|x|} ) ( 1+|x|=|x+1| ) So ( mathcal{L} text { is } 1 times geqslant 0} cup{-1} )

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