（2x+1／x）的6次方,展开第五项的二项式系数是?

y=(x^2-2x+1)／(x^2-1)
=(x-1)^2/(x+1)(x-1)
=(x-1)/(x+1)
=(x+1-2)/(x+1)
=1-2/(x+1)
显然 ,2/(x+1) 可以取到不等于0的任何实数
所以,y=(x+1)/(x-1)的值域 是y≠ 1

原式=(x-1)²/(x+1)(x-1)×(x+1)/x(x-1)-1/x+1
=1/x-1/x+1
=1

√（2x+1）在分子上,只需考虑根号里面的部分,所以2x+1>=0,得x>=-1/2,分母上1-|x|不能为0,则|x|不等于1,即x≠±1,综上可得x>=-1/2且x≠1.

(x2-2x+1／x2-1)÷﹙1+(x-3／x+1)
=(x-1)(x-1)/(x+1)(x-1) ÷﹙2(x-1)／x+1)
=(x-1)/(x+1) × (x+1)/2(x-1)
=1/2