Question

( 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} )

# 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)

Solution