a, 2017^x = 1
2017^x = 2017^0
=> x = 0
b, 2^x+3 . 2^x = 128
=> 2^x+3+x = 128
=> 2^2x+3 = 2^7
=> 2x + 3 = 7
=> 2x = 7 - 3
=> 2x = 4
=> x = 4 : 2
=> x = 2
c, x^2016 = x^2017
=> x = 0 hoặc x = 1
d, (x - 5)^2000 = (x - 5)^2002
=> x - 5 = 0 hoặc x - 5 = 1
* x - 5 = 0
x = 0 + 5
x = 5
* x - 5 = 1
x = 1 + 5
x = 6