（1）若a-b=4,b+c=2,求：a^2+b^2+c^2-ab+bc+ca的值
若a-b=4,
b+c=2
两式相加,得
a+c=6
a^2+b^2+c^2-ab+bc+ca
=(2a^2+2b^2+2c^2-2ab+2bc+2ca)/2
=(a^2-2ab+b^2+b^2+2bc+c^2+a^2+2ac+c^2)/2
=[(a-b)^2+(b+c)^2+(a+c)^2]/2
=(4^2+2^2+6^2)/2
=(16+4+36)/2
=56/2
=28
（2）已知x^2-3x-1=0,则x^2+1/x^2=
x^2-3x-1=0
当x不等于0时,
x-3-1/x=0
x-1/x=3
(x-1/x)^2=9
x^2-2*x*(1/x)+1/x^2=9
x^2-2+1/x^2=9
x^2+1/x^2=11