Question

【求解答】方程2x²-2x-1=0的两个实数根分别为x1,x2,则

(1)(x1-1)(x2-1)=_____
(2)1/x1+1/x2=_______
(3) | x1-x2 | =________
提示: | x1-x2 | =√(x1-x2)²=√(x1+x2)²-4x1x2
在线求解答

Answer 1

解：根据根与系数的关系，x1+x2=-b/a=-(-2)/2=1,x1*x2=c/a=-1/2
所以1）(x1-1)(x2-1)=x1x2-(x1+x2)+1=-1/2-1+1=-1/2
2)1/x1+1/x2=(x1+x2)/(x1x2)=1/（-1/2)=-2
3) | x1-x2 | =√(x1-x2)²=√(x1+x2)²-4x1x2=√1-4*(-1/2)=√3

Answer 2

由2x²-2x-1=0
得x²-x-1/2=0
有根与系数的关系 x1x2=-1/2 , x1+x2=1
(1)(x1-1)(x2-1)=x1x2-(x1+x2)+1=-1/2
(2)1/x1+1/x2=(x1+x2)/x1x2 =-2
(3)| x1-x2 | =√(x1-x2)²=√(x1+x2)²-4x1x2=√1-4(-1/2)=√3

Answer 3

由韦达定理得x1+x2=1,x1x2=-1/2
∴（x1-1)(x2-1)=x1x2-(x1+x2)+1=-1/2
1/x1+1/x2=x1+x2/x1x2=-2
x1-x2=根号下（x1+x2)²-4（x1x2)=根号3

Answer 4

X1+X2= -b/a=1，X1·X2=c/a=-1/2（也称韦达定理）
1）(x1-1)(x2-1)=x1x2-(x1+x2)+1=-1/2
2))1/x1+1/x2=(x1+x2)/x1x2=-2
3)| x1-x2 |=√(x1-x2)²=√(x1+x2)²-4x1x2=根号3

Answer 5

(1)负的二分之一 (2)负一加根号三 (3)二分之根号三