Question

求一个方程式的解

已知a,b,c均不为零，X1,x2是关于X方程ax平方+bx+c=0的两个实根，则1/X1+1/X2等于多少？

Answer 1

根据韦达定理得：
x1 + x2 = -b/a
x1 * x2 = c/a

1/x1 + 1/x2
= (x1 + x2)/(x1 * x2)
= (-b/a)/(c/a)
= -b/c

Answer 2

X1,x2是关于X方程ax平方+bx+c=0的两个实根
则 x1+x2=-b/a x1x2=c/a

所以
1/X1+1/X2=(x1+x2)/x1x2=-b/a /c/a=-b/c

Answer 3

-b/c

Answer 4

根据韦达定理得：
x1 + x2 = -b/a
x1 * x2 = c/a

1/x1 + 1/x2
= (x1 + x2)/(x1 * x2)
= (-b/a)/(c/a)
= -b/c

Answer 5

x1 + x2 = -b/a
x1 * x2 = c/a

1/x1 + 1/x2
= (x1 + x2)/(x1 * x2)
= (-b/a)/(c/a)
= -b/c