Question

设x1，x2是方程2x²+3x+1=0的2个根，不解方程，求下列各式的值

1.(x1+1)(x2+1) 2. x2／x1+x1／x2

Answer 1

由方程的根与系数关系有：x1+x2=-3/2,x1*x2=1/2
1.(x1+1)(x2+1)=x1*x2+x1+x2+1=1/2-3/2+1=0
2.x2/x1+x1/x2=(x1^2+x2^2)/x1*x2=[(x1+x2)^2-2x1*x2]/x1*x2=(9/4-1)/1/2=5/2

Answer 2

(x1+1)(x2+1) =x1x2+x1+x2+1=1/2+(-3/2)+1=0
x2／x1+x1／x2=[(x1+x2)^2-2x1x2]/(x1x2)=5/2

Answer 3

由原式得x1+x2=-2/3,x1*x2=1/2 (1)x1*x2+x1+x2+1=0 (2)得(x1+x2)^2-2x1x2/x1*x2=-13/6

Answer 4

式子呢？

追问

在下面