Question

已知x²-2=0,求代数式[(x-1)²]/(x²-1)+(x²)/(x+1)的值.

已知x²-2=0,求代数式[(x-1)²]/(x²-1)+(x²)/(x+1)的值.

Answer 1

x^2-2=0,
x^2=2
原式=（x^2+1-2x)/x^2-1+x^2/x+1
把x^2=2代入，得
（1-2x)/2+1/x+1
将式子通分，得
(（1-2x)(x+1)+2)/2(x+1)
化简后将x^2=2代入，化简，得
-（x+1)/2(x+1)
= -1/2

Answer 2

[(x-1)²]/(x²-1)+(x²)/(x+1)
= [(x-1)²]/(x²-1)+(x²)(x-1)/[(x+1)(x-1)]
= (x² - 2x + 1 + x³ - x²)/ (x²-1)
= [x(x²-2) + 1]/(x²-2 + 1)
= (x*0 + 1)/(0 + 1)
= 1

Answer 3

由已知得：x²=2
[(x-1)²]/(x²-1)+(x²)/(x+1)
=[(x-1)²]/(x+1)(x-1)+x²/(x+1)
=[(x-1)+x²]/(x+1)
=(x+1)/(x+1)
=1

Answer 4

x^2-2=0 ->x^2=2
->(x-1)^2/(x^2-1)+x^2/(x+1)=(x-1)^2+2*(x-1)/[(x+1)*(x-1)]=(x-1)^2+2x-2=x-2x+1+2x-2=1

Answer 5

题目有误？