For a^b=1 either a=1, b=0, or a=-1 and b is an even integer.
Therefore in the case of (x^2-7x+11)^(x^2-11x+30)=1 ...
(x^2-11x+30)=0
(x-5)(x-6)=0
x=5 or x=6 (both valid)
(x^2-7x+11)=1
x^2-7x+10=0
(x-5)(x-2)=0
x=5 or x=2 (both valid)
x^2-7x+11=-1
x^2-7x+12=0
(x-4)(x-3)=0
x=4 or x=3
Testing validity...
when x=4 (x-5)(x-6)=2 which is an even integer
when x=3 (x-5)(x-6)=6 which is an even integer
Therefore both are valid
Final solutions: x=5, 6, 2, 3 or 4
In the case of (2-x^2)^(x^2-(3√2)x+4)=1 ...
x^2-(3√2)x+4=0
Using the quadratic formula...
x=(3√2+or-√2)/2
x=2√2 or x=√2 (both valid)
2-x^2=1
x^2-1=0
(x+1)(x-1)=0
x=-1 or x=1 (both valid)
2-x^2=-1
x^2-3=0
x=+or-√3
Testing validity...
when x=+or-√3 x^2-(3√2)x+4 does not equal an integer
Therefore both are invalid
Final solutions: x=2√2, √2, -1, or 1