$Thesumofallrealvaluesofxsatisfyingtheequation{(x^2-5x+5)}^{x^2+4x-60}=1$

$\begin{array}{l}Given{(x^2-5x+5)}^{x^2+4x-60}=1\\ Case\left(i\right):x^2+4x-60=0\\ \Rightarrow (x+10)(x-6)=0\Rightarrow x=-10,6\\ Case\left(ii\right):x^2-5x+5=1\\ \Rightarrow x^2-5x+4=0\\ \Rightarrow (x-1)(x-4)=0\Rightarrow x=1,4\\ Case\left(iii\right):x^2-5x+5=-1\\ \Rightarrow x^2-5x+6=0\\ \Rightarrow (x-2)(x-3)=0\Rightarrow x=2,3\\ Ifx=3then{x}^2+4x-60isoddnumber\\ \therefore x=2isonlyacceptable\\ \therefore sumof\mathrm{int}egers=-10+4+1+6+2=3\end{array}$