Question

已知a=√2-b+√b-2+3,求√a-√b/√a+√b-√a+√b/√a-√b的值

Answer 1

a=√(2-b)+√(b-2)+3
因为√(2-b)>=0,√(b-2)>=0
即b<=2,b>=2
所以b=2,a=3

(√a-√b)/(√a+√b)-(√a+√b)/(√a-√b)
=(√a-√b)^2/(√a+√b)(√a-√b)-(√a+√b)^2/(√a+√b)(√a-√b)
=(√a-√b)^2/(a-b)-(√a+√b)^2/(a-b)
=(a+b-2√ab)/(a-b)-(a+b+2√ab)/(a-b)
=(a+b-2√ab-a-b-2√ab)/(a-b)
=(-4√ab)/(a-b)
=-4√(2*3)/(3-2)
=-4√6

Answer 2

a=√2-b+√b-2+3
所以 2-b=b-2=0
b=2
a=3
√a-√b/√a+√b-√a+√b/√a-√b
=[(√a-√b)^2-(√a+√b)^2]/(√a+√b)(√a-√b)
=-4√ab/(a-b)
=-4√2*3/(3-2)
=-4√6