记方程（1）x^2-3x+2=0的两根之和为a1 (2) x^2+7x+12=0两根之和为a2 (3)x^2-11x+30=0两根之和为a3 求a2006记方程（1）x^2-3x+2=0的两根之和为a1 (2) x^2+7x+12=0两根之和为a2 (3)x^2-11x+30=0两根之
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记方程（1）x^2-3x+2=0的两根之和为a1 (2) x^2+7x+12=0两根之和为a2 (3)x^2-11x+30=0两根之和为a3 求a2006记方程（1）x^2-3x+2=0的两根之和为a1 (2) x^2+7x+12=0两根之和为a2 (3)x^2-11x+30=0两根之
记方程（1）x^2-3x+2=0的两根之和为a1 (2) x^2+7x+12=0两根之和为a2 (3)x^2-11x+30=0两根之和为a3 求a2006
记方程（1）x^2-3x+2=0的两根之和为a1
(2) x^2+7x+12=0两根之和为a2
(3)x^2-11x+30=0两根之和为a3
求a2006的值
求a1+a2+a3……+a2008
记方程（1）x^2-3x+2=0的两根之和为a1 (2) x^2+7x+12=0两根之和为a2 (3)x^2-11x+30=0两根之和为a3 求a2006记方程（1）x^2-3x+2=0的两根之和为a1 (2) x^2+7x+12=0两根之和为a2 (3)x^2-11x+30=0两根之
由韦达定理 x1+x2=-b/a
∴ a1=3,a2=-7,a3=11
有 |a2|-|a1|=|a3|-|a2|=4
|an|=4n-1
an=(4n-1)*(-1)^(n+1),n∈N*
∴ a2006=(4*2006-1)(-1)^(2006+1)= -8023
S2008=(a1+a2)+(a3+a4)+……+(a2007+a2008)
=(-4)*2008/2= -4016
x1，x2是方程的根，由韦达定理得x1+x2=3 x1代入方程，得x1 -3x1+1=0 x1 =3x1-1 4x1 +12x2+11 =4(3x1-1)+12x2+11 =12x1+12