解:(1)∵x

,x

是方程x

-6x+k=0的两个根
∴x

+ x

=6 x

x

=k······················1分
∵x

x

—x

—x

=115
∴k

—6=115·············································2分
解得k

=11,k

=-11······································3分
当k

=11时

=36—4k=36—44<0 ,∴k

=11不合题意·······4分
当k

=-11时

=36—4k=36+44>0∴k

=-11符合题意·········5分
∴k的值为—11············································6分
(2)x

+x

=6,x

x

=-11·····························7分
而x


+x


+8=(x

+x

)

—2x

x

+8=36+2×11+8=66·····9分
(1)方程有两个实数根,必须满足△=b
2-4ac≥0,从而求出实数

的取值范围,再利用根与系数的关系,
1
2-
1-
2=115.即
1
2-(
1+
2)=115,即可得到关于

的方程,求出

的值.
(2)根据(1)即可求得
1+
2与
1
2的值,而
12+
22+8=(
1+
2)
2-2
1
2+8即可求得式子的值