先化简 再求值:已知x的平方-2=0,求代数式(x-1)的平方除以(x的平方-1)+x的平方⼀(x+1)的值

(x-1)²/(x²-1)+x²/(x+1)
=[(x-1)²+x²(x-1)]/(x²-1)
=(x²-2x+1+x³-x²)/(x²-1)
=(-2x+1+x³)/(x²-1)
=[1+x(x²-2)]/(x²-2+1)
=1/1
=1

[(a²-b²)/(a²-ab)]/[a+(2ab+b²)/a]
=[(a²-b²)/(a²-ab)]/[(a²+2ab+b²)/a]
=[(a²-b²)/(a²-ab)]*[a/(a+b)²]
=(a²-b²)(a+b)²/(a-b)
=(a+b)³

代数式(x-1)的平方除以(x的平方-1)+x的平方/(x+1)
=(x-1)(x-1)/(x-1)(x+1)+x^2/(x+1)
=(x-1)/(x+1)+x^2/(x+1)
=(x^2+x-1)/(x+1)
x^2-2=0即x^2=2
代入得(x^2+x-1)/(x+1)
=(2+x-1)/(x+1)
=（x+1)/(x+1)
=1
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（a的平方-b的平方）/（a的平方-ab）/

=(a+b)/a+....看不懂题目？

(x-1)的平方/(x的平方-1)+x的平方/(x+1)
=(x-1)的平方/[(x-1)(x+1)]+x的平方/(x+1)
=(x-1)/(x+1)+x的平方/(x+1)
=(x的平方+x-1)/(x+1)
x的平方=2
原式=(2+x-1)/(x+1)
=(x+1)/(x+1)
=1
（a的平方-b的平方）/（a的平方-ab）/[a+（2ab+b的平方）/a]
=(a-b)(a+b)/[a(a-b)]/[(a+b)的平方/a]
=(a+b)/a*a/(a+b)的平方
=1/(a+b)

解：﹙x－1﹚²/﹙x²－1﹚＋x²/﹙x＋1﹚
＝﹙x－1﹚²/﹙x－1﹚﹙x＋1﹚ ＋x²/﹙x＋1﹚
＝﹙x－1﹚/﹙x＋1﹚－2/﹙x＋1﹙
＝﹙x＋1﹚/﹙x＋1﹚
＝1