Question

求解 初中数学 分式问题

Answer 1

解：
x²-x-1=0 x²=x+1
(x⁴+2x+1)/x^5
=[(x²)²+2x+1]/[x(x²)²]
=[(x+1)²+2x+1]/[x(x+1)²]
=(x²+4x+2)/[x(x²+2x+1)]
=(x+1+4x+2)/[x(x+1+2x+1)]
=(5x+3)/[x(3x+2)]
=(5x+3)/(3x²+2x)
=(5x+3)/[3(x+1)+2x]
=(5x+3)/(5x+3)
=1
提示：就是反复用x+1代替x²

Answer 2

答：

x^2-x-1=0
则x^2=x+1
所以：
(x^4+2x+1)/x^5

=[(x+1)^2+2x+1] /[x(x+1)^2]
=(x^2+4x+2) / (x^3+2x^2+x)
=(x^2+4x+2)/[(x+1)x+2(x+1)+x]
=(x^2+4x+2)/(x^2+4x+2)
=1

Answer 3

x`2=x+1,直接代进去，x`5=x`2*x`2*x=（x+1)（x+1)x=(x`2+2x+1)x=(3x+2)x=5x+3,x`4+2x+1=(x+1)`2+2x+1=5x+3,最后结果是1

Answer 4

由已知x2=x+1，所以x4+x+x+1=x4+x+x2=x4+x*（x+1）=x4+x*x2=x4+x3=x3*(x+1)=x3*x2=x5，
（x4+2x+1）/x5=1

Answer 5

x²=x+1
原式=【（x+1）²+2x+1】/【x（x+1）²】=【x²+2x+1+2x+1】/【x³+2x²+x】
=【x²+4x+2】/【x（x+1）+2（x+1）+x】=【x+1+4x+2】/【x²+x+2x+2+x】
=【5x+3】/【x²+4x+2】=【5x+3】/【x+1+4x+2】=【5x+3】/【5x+3】=1