已知:x2-3x+1=0,计算下列各式的值: (1)x2+1/x2+2; (2)2x3-3x2-7x+2009.

问题描述:

已知:x2-3x+1=0,计算下列各式的值:
(1)x2+

1
x2
+2;
(2)2x3-3x2-7x+2009.

(1)∵x2-3x+1=0,
∴x≠0,
∴x-3+

1
x
=0,即x-
1
x
=3,
∴(x-
1
x
2=9,
∴x2-2+
1
x2
=9,即x2+
1
x2
=11,
∴x2+
1
x2
+2=11+2=13;
(2)∵x2-3x+1=0,
∴x2=3x-1,
∴2x3-3x2-7x+2009=2x(3x-1)-3(3x-1)-7x+2009
=6x2-18x+2012
=6(3x-1)-18x+2012
=2006.