a=\frac{3\sqrt3-4}{1+2\sqrt3}*\frac{1-2\sqrt3}{1-2\sqrt3}=
=\frac{3\sqrt3-18-4+8\sqrt3}{1^2-(2\sqrt3)^2}=
=\frac{11\sqrt3-22}{1-12}=\frac{11\sqrt3-22}{-11}=2-\sqrt3
i
b=\sqrt{27}*\frac{(\frac{1}{9})^3}{3^{-5}}=
=3\sqrt3*((\frac{1}{3})^6:3^{-5})=
=3\sqrt3(3^{-6}:3^{-5})=
=3*3^{\frac{1}{2}}*3^{-1}=3^{1+\frac{1}{2}-1}=3^{\frac{1}{2}}
c)
a+b=2-\sqrt3+3^{\frac{1}{2}}=2-\sqrt3+\sqrt3=2
80% stanowi 2
100% stanowi c
80%c=200%
c=2,5