第90页 - 亮点给力提优课时作业本八年级数学江苏版

C

2

$ -\frac {1}{3}$

解：原式$=[(\frac {1}{a-1})²- (\frac {1}{a+1})^2]÷( \frac {1}{a-1}+ \frac {1}{a+1}) · \frac {(a-1)(a+2)}{(a-2)(a+2)}$

$=(\frac {1}{a-1}-\frac {1}{a+1}) · \frac {a-1}{a-2}=\frac {2}{(a+1)(a-1)} · \frac {a-1}{a-2}=\frac {2}{a²-a-2}$

∵$2a²-2a-3=0$，∴$2a²-2a=3$，即$a²-a=\frac {3}{2}$

∴原式$= \frac 2{\frac {3}{2}-2}=-4$

C

1

$\frac {1}{2}$

解：由题意，得$a≠0$，$b≠0$

∵$2023b=\frac {1}{b}+2024$，∴$\frac {1}{b}=2023b-2024$，$2023b²-2024b-1=0$

∵$2023a²-2024a-1=0$，∴$2023(a²-b²)-2024(a-b)=0$，即$(a-b) · [2023(a+b)-2024]=0$

∵$a≠b$，∴$a-b≠0$，即$2023(a+b)-2024=0$，∴$2023(a+b)=2024$

∵$ 2023a²-2024a-1=0$，∴$\frac {1}{a}=2023a-2024$

∴$\frac {1}{a}+\frac {1}{b}=2023a-2024+2023b-2024=2023(a+b)-2024×2=2024-2024×2=-2024$