a) $A=5\left( x+3 \right)\left( x-3 \right)+{{\left( 2x+3 \right)}^{2}}+{{\left( x-6 \right)}^{2}}$
$A=5\left( {{x}^{2}}-9 \right)+\left( 4{{x}^{2}}+12x+9 \right)+\left( {{x}^{2}}-12x+36 \right)$
$A=5{{x}^{2}}-45+4{{x}^{2}}+12x+9+{{x}^{2}}-12x+36$
$A=10{{x}^{2}}$
b) khi $x=\frac{-1}{5}$ $\Rightarrow A=10.{{\left( \frac{-1}{5} \right)}^{2}}=10.\frac{1}{25}=\frac{2}{5}$