Question

已知a≠0.化简根号下4-（a+a分之一）^2-根号下4+（a-a分之一）^2

Answer 1

√[4-(a+1/a)^2]-√[4+(a-1/a)^2]
=√[4-a^2-2-1/a^2]-√[4+a^2-2+1/a^2]
=√[-a^2+2-1/a^2]-√[a^2+2+1/a^2]
=√[-(a^2-2+1/a^2)]-√[a^2+2+1/a^2]
=√[-(a-1/a)^2]-√[(a+1/a)^2]
-(a-1/a)^2>=0
(a-1/a)^2=0
a-1/a=0
a^2-1=0
a=±1

当a=1时
原式=√[-(a-1/a)^2]-√[(a+1/a)^2]
=√[-(1-1/1)^2]-√[(1+1/1)^2]
=-√4
=-2

当a=-1时
原式=√[-(a-1/a)^2]-√[(a+1/a)^2]
=√[-(1-1/1)^2]-√[(-1-1/1)^2]
=-√4
=-2