(1/√3)^2x+6*(1/√3)^x-27=0
(1/√3)^x=t>0 ⇒
t²+6t-27=0 D=144
t₁=3 x₂=-9∉
(1/√3)^x=3
3^(-x/2)=3^1
-x/2=1
x=-2.
Ответ: х=-2.
((4-b²)/(4b+2b²)-1/(2b))/((b-1)/b³)=((2-b)(2+b)/(2b*(2+b))/((b-1)/b³)=
=((2-b)/(2b)-1/(2b))/((b-1)/b³)=((2-b-1)/(2b))/((b-1)/b³=((1-b)/(2b))/((b-1)/b³=
=(1-b)*b³/((b-1)*2b)=-b²/2=-(√5)²/2=-5/2=-2,5.