Question

Срочносрочносрочносрочносрочносрочносрочно

Answer 1

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

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