1.已知a、b、c为实数,且abc=1,则(1/a+ab+1)+(1/b+bc+1)+(1/c+ca+1)d的值为（）A.2B.1C.3D.42.已知xy=-12,x+y=-4,则（x+1/y+1）+（y+1/x+1）=?3.先化简,再求值：（（x-1/x）-（x-2/x+1））÷（2x²-x/x²+2x+1）,其中

Question

1.已知a、b、c为实数,且abc=1,则(1/a+ab+1)+(1/b+bc+1)+(1/c+ca+1)d的值为（ ）A.2 B.1 C.3 D.4
2.已知xy=-12,x+y=-4,则（x+1/y+1）+（y+1/x+1）=?
3.先化简,再求值：（（x-1/x）-（x-2/x+1））÷（2x²-x/x²+2x+1）,其中x满足x²-x-1=0.

Answer 1

1.B
2.-34/15
3.（(x-1)/x-(x-2)/(x+1）)÷(2x² –x)/(x² +2x+1)
=(x²-1-x²+2x)/[x(x+1)]×(x+1)²/x(2x-1)
=(2x-1)(x+1)²/[x²(x+1)(2x-1)]
=(x+1)/x²
因为
x²-x-1=0
即
x²=x+1
所以
原式=x²/x²=1