第83页 - 江苏版启东中学作业本八年级数学（上下册）

B

D

$\frac {2}{3}$

$-\frac {1}{x}$

$±\frac {1}{2}$

解：原式$= \frac {3x^2+3xy+x^2-xy}{(x-y)(x+y)}· \frac {(x-y)(x+y)}{x}=\frac {2x(2x+y)}{(x-y)(x+y)}· \frac {(x-y)(x+y)}{x}=2(2x+y)$

∵$2x+y-3=0，$∴$2x+y=3$

∴原式$=2×3=6$

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

$=\frac 3{a-2}· \frac {(a-2)^2}{a(a-2)}=\frac 3{a}$

$ $当$a=0，$$a=2$时，原式没有意义

∴当$a=2025$时，原式$=\frac 3{2025}=\frac 1{675}$

解：$(1)$由$a+b=ab，$得到$a=b(a-1)，$即$b=\frac {a}{a-1}$

当$a=2$时，$b=2；$当$a=3$时，$b=\frac 32$

$(2)$若$a=2，$$b=2，$$\frac {b}a+\frac {a}b=1+1=2，$$ab=4$

则$\frac {b}a+\frac {a}b$比$ab{小}2$

$(3)$将$b=\frac {a}{a-1}$代入$\frac {b}{a}+\frac {a}{b}，$

得$\frac {\frac {a}{a-1}}{a}+\frac {a}{\frac a{a-1}}=\frac 1{a-1}+ a-1$

将$b=\frac {a}{a-1}$代入$ab，$得$a· \frac {a}{a-1}=\frac {a^2}{a-1}$

∵$ab-(\frac {b}{a}+\frac {a}{b})=\frac {a^2}{a-1}-\frac 1{a-1}-a+1=\frac {a^2-1}{a-1}-a+1=a+1-a+1=2$

∴$\frac {b}{a}+\frac {a}{b}$比$ab{小}2$