题目

求这三道题的答案（分式、反比例函数）

提问时间：2020-09-30

答案

1.原式=(x²/x+1)÷(x²-2/x²-1)=[x²(x-1)]/(x²-2)
代入x=根号2+1得原式=【(3+2倍根号2)*根号2】/(1+2倍根号2)
=(8+5倍根号2)/7.
2.原式=[(x²-x)/(x-2)]÷[(x-1)²/(x-2)]
=x/(x-1)
代入得原式=(根号2+1)/根号2
=(2+根号2)/2.
3.原式=3-1/(a+1)-1/(b+1)-1/(c+1)
=3-(y+z)/(x+y+z)-(x+z)/(x+y+z)-(x+y)/(x+y+z)
=3-2=1.
希望对你有帮助.