Question

Помогите решить уравнение

Answer 1

$[frac{sqrt{b}+3}{sqrt{b}-3}+frac{sqrt{b}-3}{sqrt{b}+3}-frac{36}{b-9}]^{-2} \ \ 1)frac{sqrt{b}+3}{sqrt{b}-3}+frac{sqrt{b}-3}{sqrt{b}+3}=frac{(sqrt{b}+3)^2+(sqrt{b}-3)^2}{(sqrt{b}-3)(sqrt{b}+3)}=frac{(b+3sqrt{b}+3sqrt{b}+9)+(b-3sqrt{b}-3sqrt{b}+9)}{(b+3sqrt{b}-3sqrt{b}-9)}= \ \ =frac{2b+18}{b-9}=frac{2(b+9)}{b-9} \ \ 2)frac{2b+18}{b-9}-frac{36}{b-9}=frac{2b+18-36}{b-9}=frac{2b-18}{b-9}=frac{2(b-9)}{b-9}=2 \ \ 3)2^{-2}=( frac{1}{2} )^2= frac{1}{4}$

Answer 2

[(√b+3)/(√b-3) +(√b-3)/(b√+3) - 36/(b-9)]^-2=
[(b+6√b+9+b-6√b+9-36)/(b-9)]^-2=[(2b-18)/(b-9)]^-2=[2(b-9)/(b-9)]^-2=(2)^-2=1/4