已知a=根号2+1,b=根号2-1,求根号ab*(根号b分之a-根号a分之b)的值
问题描述：
已知a=根号2+1,b=根号2-1,求根号ab*(根号b分之a-根号a分之b)的值
答
ab=(√2)²-1²=1
√(a/b)-√(b/a)
=√a/√b-√b/√a
=(a-b)/√ab
=(√2+1-√2+1)/√1
=2
所以原式=1×2=2
答
a=根号2+1,b=根号2-1
ab=1
根号ab*(根号b分之a-根号a分之b)
=根号b分之a-根号a分之b
=(a-b)/根号ab
=a-b
=根号2+1-(根号2-1)
=2