已知,x눀-2=0,求代数式(x-1)눀 x눀 ------------ + __________ 的值 x눀-1 x+1

已知,x^2-2=0,求代数式(x-1)^2⼀x^2-1 + x^2⼀x+1的值
2024-12-20 18:13:53
推荐回答(2个)
回答1:

x²-2=0,所以x²=2,x=±根号2

代数式变形:
(x-1)²/(x-1)(x+1) +x²(x-1)/(x-1)(x+1)
=[(x-1)²+x²(x-1)] / (x-1)(x+1)
提取(x-1)
(x-1) [(x-1)+x²] /(x-1)(x+1)
消去(x-1)
[(x-1)+x²] / (x+1)
x²+x-1 / (x+1)
x(x+1)-1 / (x+1)
然后将±根号2代入
=根号2(根号2+1)-1/根号2+1
=2+根号2-1/根号2+1
=1根号2+1/根号2+1
=1
或者=-根号2(-根号2+1)-1/-根号2+1
=2-根号2-1/-根号2+1
=-根号2+1/-根号2+1
=1
综上就是1

回答2:

问题未描述清楚,请列明详细问题!要求的是哪个数式?