[(x-1)(x+4)]²+(x+1)²=(x-3)²+[(x-1)(x-2)]²

作差：
[(x-1)(x+4)]²+(x+1)²-(x-3)²+[(x-1)(x-2)]²
=(x-1)^2[(x+4)^2-(x-2)^2]+[(x+1)²-(x-3)²]
=12(x-1)^2(x+1)+8(x-1)
=4(x-1)(3(x^2-1)+2)
=4(x-1)(3x^2-1)
=0
x=1或±根号3/3

[(x-1)(x+4)]^2+(x+1)^2=(x-3)^2+[(x-1)(x-2)]^2
[(x-1)(x+4)]^2-[(x-1)(x-2)]^2=(x-3)^2-(x+1)^2
[(x-1)(x+4)-(x-1)(x-2)].[(x-1)(x+4)+(x-1)(x-2)]=[ (x-3)-(x+1)].[ (x-3)+(x+1)]
12(x-1)(x^2-1) =-8(x-1)
3(x-1)(x^2-1) =-2(x-1)
(x-1)[ 3(x^2-1) +2 ] =0
(x-1).(3x^2-1)=0
x=1 or √3/3 or - √3/3