以叙述,写出下式:
(ab)^2 + a^2 + b^2 + 1 = 4ab
移项,整理得
(ab)^2 - 2ab + 1 + a^2 - 2ab + b^2 = 0
(ab - 1)^2 + (a - b)^2 = 0
所以ab - 1 = 0,a = b
所以a = b = 1 或a = b = -1
以叙述,写出下式:
(ab)^2 + a^2 + b^2 + 1 = 4ab
移项,整理得
(ab)^2 - 2ab + 1 + a^2 - 2ab + b^2 = 0
(ab - 1)^2 + (a - b)^2 = 0
所以ab - 1 = 0,a = b
所以a = b = 1 或a = b = -1