若lgx^2=lg(根号2-1)-lg(根号2+1),则x+1/x＝?

lg(√2-1)-lg(√2+1),
=lg[(√2-1)/(√2+1)]
=lg(√2-1)^2
=lgx^2
所以x^2=(√2-1)^2=3-2√2
所以x=√2-1或1-√2
x+1/x=(x^2+1)/x=(3-2√2+1)/x=(4-2√2)/x
x=√2-1
则x+1/x=(4-2√2)(√2+1)/1=2√2
x=1-√2
则x+1/x=(4-2√2)(√2+1)/(-1)=-2√2

是lg（x^2）？
lg(√2-1)-lg(√2+1)，
=lg[(√2-1)/(√2+1)]
=lg(√2-1)^2
=lgx^2
所以x^2=(√2-1)^2
所以x=√2-1或1-√2
x=√2-1
则x+1/x=2√2
x=1-√2
则x+1/x=-2√2

2√2

lgx^2=lg(根号2-1)-lg(根号2+1)，
lgx^2=lg[(根号2-1)/(根号2+1)]
lgx^2=lg(根号2-1)^2
x=根号2-1
x+1/x＝2根号2