一、求值例 1　 ( 1995年全国初中数学联赛试题 )已知ａ、ｂ是方程ｘ2 +(ｍ-2 )ｘ +1=0的两根 ,则 ( 1+ｍａ +ａ2 ) ( 1+ｍｂ +ｂ2 )的值是(　　 )(Ａ) 1　 (Ｂ) 2　 (Ｃ) 3　 (Ｄ) 4分析　由题意和韦达定理知 :ａ +ｂ=2 -ｍ ,ａｂ=1.ａ2 +ｍａ +1=ａ2 +(ｍ -2 )ａ+1+2ａ　　　 First, the evaluation example 1 (1995 national junior high school mathematics league questions) is known a, b is the two equations x2 + (m-2) x +1 = 0, then (1+ma + a2) (1+mb The value of +b2) is ()(A) 1 (B) 2 (C) 3 (D) 4 The analysis is based on the problem meaning and the Vedic theorem: a +b = 2 -m, ab = 1.a2 +ma + 1=a2 +(m -2 )a+1+2a