已知x=1⼀2(根号下a⼀b+根号下b⼀a),(a>b>0),求2倍根号下ab⼀x-根号下x方-1的值

2025-05-14 11:05:41
推荐回答(2个)
回答(1):

a>b>0
x=1/2[根号(a/b)+根号(b/a)]=1/2{根号[b^2/(ab)]+根号[b^2/(ab)]}=(a+b)/[2根号(ab)]

2根号(ab)/x)-根号(x^2-1)
=2根号(ab)*2根号(ab)/(a+b)-根号{(a+b)^2/(4ab)-1}
=4ab/(a+b)-根号[(a-b)^2/(4ab)]
=4ab/(a+b)-(a-b)/[2根号(ab)]
=[8a^2b^2-(a^2-b^2)根号(ab)]/[2ab(a+b)]

回答(2):

1