Question

化简求值:（x分之x-1 减x+1分之x-2）除以 x的平方+2x+1分之2x的平方-x. 其中x满足x的平方-x-1=0

Answer 1

因为X满足x平方-x-1=0所以2分之x-1的平方平方=0所以x=1，你把1带进去就行了，我也不知道对不对，你在考虑一下
x^2-x-1=0
x^2=x+1

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

Answer 2

x^2-x-1=0
x^2=x+1

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

Answer 3

太难了

Answer 4

x²-x-1=0
所以x²-1=x,x²=x+1

[(x-1)/x-(x-2)/(x+1)]/[(2x²-x)/(x²+2x+1)]
=[(x-1)/(x²-1)-(x-2)/(x+1)]*[(x+1)²/(2x²-x)]
=[1/(x+1)-(x-2)/(x+1)]*[(x+1)²/(2x²-x)]
=[(3-x)/(x+1)]*[(x+1)²/(2x²-x)]
=(3-x)(x+1)/(2x²-x)
=(2x+3-x²)/(2x²-x)
=(2x+3-x-1)/(2x+2-x)
=(x+2)/(x+2)
=1