先化简，再求值：{x-[x⼀(x+1)]}除以{1+[1⼀(x^2-1)]}，其中x=根号2+1

{x-[x/(x+1)]}除以{1+[1/(x^2-1)]}
=【(x²+x-x)/(x+1)】/[(x²-1+1)/(x+1)(x-1)]
=【x²/(x+1)】/[(x²)/(x+1)(x-1)]
=x-1
=根号2+1-1
=根号2

{x-[x/(x+1)]}÷{1+[1/(x^2-1)]}
={x(x+1)/(x+1)-x/(x+1)}÷{(x^2-1)/(x^2-1)+1/(x^2-1)}
={[x(x+1)-x]/(x+1)}÷{(x^2-1+1)/(x^2-1)}
={[x^2+x)-x]/(x+1)}÷{(x^2-1+1)/(x^2-1)}
=x^2/(x+1)÷{x^2/(x^2-1)}
=x^2/(x+1)*(x^2-1)/x^2
=(x^2-1)/(x+1)
=(x-1)(x+1)/(x+1)
=x-1
=√2+1-1
= √2

解：原式=x(1-1/(x+1))/(x^2/(x^2-1))=(x/(x+1))/(x/[(x+1)(x-1)])=x-1
当x=1+√2时，原式=1+√2-1=√2

{x-[x/(x+1)]}除以{1+[1/(x^2-1)]}
原式=[x^2+x-x/(x+1)]除以[x^2-1+1/(x^2-1)]
=[x^2/(x+1)]除以[x^2/(x^2-1)] (约分)
=1/(x-1)
代入x=得
原式=1/(根号2+1-1)
=1/根号2
=根号2/2