Question

已知x^2-6x+4的根是a，b.求√a-√b/√a+√b的值

Answer 1

已知x^2-6x+4=0的根是a，b
根据韦达定理得
a+b=6
ab=4

(a-b)^2
=(a+b)^2-4ab
=6^2-4*4
=36-16
=20
a-b=±2√5

(√a-√b)/(√a+√b)
=(√a-√b)^2/[(√a+√b)(√a-√b)]
=(a+b-2√ab)/(a-b)
=(6-2√4)/(a-b)
=(6-4)/(a-b)
=2/(a-b)
=2/(±2√5)
=±1/√5
=±√5/5

Answer 2

a+b=6,ab=4
(a-b)^2=(a+b)^2-4ab=20
因为a>b,所以a-b=2根号下5

(根号下a-根号下b)/(根号下a+根号下b)
=(根号下a-根号下b)^2/(根号下a+根号下b)(根号下a-根号下b)
=(a+b-2根号下ab)/(a-b)
=（6-2根号下4）/2根号下5
=2/2根号下5
=(根号下5)/5