1.a-b=2+√3,b-c=2-√3
相加
a-c=4
a^2+b^2+c^2-ab-ac-bc
=(2a^2+2b^2+2c^2-2ab-2bc-2ca)/2
=[(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2)]/2
=[(a-b)^2+(b-c)^2+(a-c)^2]/2
=[(2+√3)^2+(2-√3)^2+4^2]/2
=(4+4√3+3+4-4√3+3+16)/2
=30/2
=15
2.a4+2a3-3a2-4a+3
=a2(a2+a)+a3-3a2-4a+3
=-a2+a3-3a2-4a+3
=a3-4a2-4a+3
=a3-4(a2+a)+3
=a3+7
由已知：a2+a=-1 得a3+a2=-a 所以a3=-a2-a=-(a2+a)=1
所以原式=8
3.2/(x-2)+mx/(x^2-4)=3/(x+2)
2(x+2)+mx=3(x-2)
(m-1)x=-10
x=2为增根时 m=-4
x=-2为增根时 m=6