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$(1-b)^2( \frac{1}{b-1^2} - \frac{1}{1-b^2} )+ \frac{3+b}{1+b} = \\ (1-b)^2( \frac{1}{b-1} - \frac{1}{(1-b)(1+b)} )+ \frac{3+b}{1+b} = \\ (1-b)^2( \frac{1}{b-1^2} - \frac{1}{1-b^2} )+ \frac{3+b}{1+b} = \\ \frac{(1-b)^2}{b-1} - \frac{(1-b)^2}{(1-b)(1+b)}+ \frac{3+b}{1+b} = \\ -\frac{(1-b)^2}{1-b} - \frac{1-b}{1+b}+ \frac{3+b}{1+b} = -(1-b) -( \frac{1-b}{1+b}- \frac{3+b}{1+b}) = \\ -1+b - \frac{1-b-(3+b)}{1+b} = -1+b - \frac{1-b-3-b}{1+b} = -1+b - \frac{(-2)}{1+b} = \\ -1+b + \frac{2}{1+b} =$
$\frac{-1(1+b)+b(1+b)+2}{1+b} = \frac{-1-b+b+b^2+2}{1+b} =\frac{b^2+1}{1+b} =$

При b=0,2
$\frac{b^2+1}{1+b}$ =(0,2²+1)/(1+0,2)=(0,04+1)/1,2=1,04/1,2=104/120=26/30=13/15