Question

已知关于x的一元二次方程5x²-4x-1＝0的两个解为x1和x2，求1/x1+1/x2和x2/x1+x1/x2的值

Answer 1

x1+x2=4/5
x1x2=-1/5

所以1/x1+1/x2
=(x1+x2)/(x1x2)
=-4

(x1+x2)²=(4/5)²
x1²+2x1x2+x2²=16/25
x1²+x2²=16/25-2x1x2=26/25
所以x2/x1+x1/x2
=(x2²+x1²)/(x1x2)
=-26/5

追问

为什么？
所以1/x1+1/x2
=(x1+x2)/(x1x2)
=-4

追答

通分都不懂？？？

Answer 2

1/x1+1/x2=(x1+x2)/x1x2
x2/x1+x1/x2=(x12+x22)/x1x2=((x1+x2)2-2x1x2)/x1x2
x1x2=-1/5 x1+x2=4/5代入即可

Answer 3

原式=（5x+1）（x-1）=0，所以x1=-1/5，x2=1
因此第一个式子=-4，第二个=-26/5

Answer 4

答案是－4和－26/5 具体方法是： 解方程得：x1=1，x2=-1/5；再将它们代人即可

Answer 5

X1，X2会求吧，
解由已知方程得：（5X1+1)(X2-1)=0
得：X1=-1/5 X2=1/2
1/X1+1/X2=-5+2=-3
X2/X1+X1/X2=-5/2+(-2/5)=-25/10-4/10=-29/10=2.9

Answer 6

韦达定理：
X1+X2=-(-4/5)=4/5;
X1X2=-1/5;

1/x1+1/x2
=(X2+X1)/(X1X2)
=(4/5)]/(-1/5)
=-4;
x2/x1+x1/x2
=(X2²+X1²)/(X1X2)
=[(X2+X1)²-2X1X2]/(X1X2)
=(X2+X1)²/(X1X2)-2
=(4/5)²/(-1/5)-2
=-16/5-2
=-26/5