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+
+2;1 x2
(2)2x3-3x2-7x+2009.
答
(1)∵x2-3x+1=0,
∴x≠0,
∴x-3+
=0,即x-1 x
=3,1 x
∴(x-
)2=9,1 x
∴x2-2+
=9,即x2+1 x2
=11,1 x2
∴x2+
+2=11+2=13;1 x2
(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.