(-1)^0 =1
1^0=1
Hence,
Either, x^2-5x+5=-1 => x^2-5 x+6 =0=> (x-2)(x-3)=0=> x=2, 3
Or, x^2--11x+30=0=0=> (x-5)(x-6) = 0=> x=5,6
Or, x^2--5x+5= 1=> x^2 -5x +4 = 0=> (x-1)(x-4)=0=> x= 1, 4
Therefore, x ε{1,2,3,4,5,6}
Consequently, precisely six distinct integers satisfy the equation.
Edited on January 18, 2024, 6:50 am