Question

已知x+2=1/x,试求代数式1/(x+1)-(x+3)/(x的平方-1)×(x的平方-2x+1)/(x的平方+4x+3)的值。

分式方程。

Answer 1

您好：

x+2=1/x
x²+2x=1

1/(x+1)-(x+3)/(x的平方-1)×(x的平方-2x+1)/(x的平方+4x+3)
=1/(x+1)-(x+3)/(x+1)(x-1) x (x-1)²/（x+1）（x+3）
=1/(x+1)-(x-1)/(x+1) ²
=(x+1-x+1)/(x+1)²
=2/(x+1)²
=2/(x²+2x+1)
=2/(1+1)
=1

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Answer 2

x+2=1/x
两边乘x
x²+2x=1

原式=1/(x+1)-(x+3)/(x+1)(x-1)×(x-1)²/(x+1)(x+3)
=1/(x+1)-(x-1)/(x+1)²
=(x+1-x+1)/(x+1)²
=2/(x²+2x+1)
=2/(1+1)
=1