Question

Помогите, пожалуйста!

Answer 1

$b^2-3b+2=0, b_1=1, b_2=2 to b^2-3b+2=(b-1)(b-2)\b^2+3b+2=0, b_1=-1, b_2=-2, to b^2+3b+2=(b+1)(b+2)\.......=frac{(b-1)(b-2)-(b-1)sqrt{(b-2)(b+2)}}{(b+1)(b+2)-(b+1)sqrt{(b-2)(b+2)}}cdot sqrt{frac{b+2}{b-2}}=\=frac{(b-1)sqrt{b-2}(sqrt{b-2}-sqrt{b+2})}{(b+1)sqrt{b+2}(sqrt{b +2}-sqrt{b-2})}cdot frac{sqrt{b+2}}{sqrt{b-2}}=frac{(b-1)(sqrt{b-2}-sqrt{b+2})}{-(b+1)(sqrt{b-2}-sqrt{b+2})}=-frac{b-1}{b+1}$