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+1x=0,即x-1x=3,∴(x-1x)2=9,∴x2-2+1x2=9,即x2+1x2=11,∴x2+1x2+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...