第40页 - 通城学典课时作业本八年级数学人教版南通专版

不变

相加减

$\frac {a}{c}±\frac {b}{c}=\frac {a±b}{c}$

通分

$\frac {a}{b}±\frac {c}{d}=$

$\frac {ad}{bd}±\frac {bc}{bd}=\frac {ad±bc}{bd}$

D

A

$\frac {3y+2x}{6x²y²}$

$\frac {a-2}{a} $

$解：原式=\frac {5a+3b-2a}{a^2-b^2}$

$=\frac {3a+3b}{(a-b)(a+b)}$

$ =\frac {3(a+b)}{(a-b)(a+b )} $

$=\frac {3}{a-b}$

$解：原式=\frac {a(a+1)}{(a+1)(a-1) }-\frac {3a-1}{(a+1)(a-1)}$

$=\frac {a^2+a-3a+1}{(a+1)(a-1)}$

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

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

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

解：原式$=\frac {(m+2)(m-2)}{m-2}-\frac {\ \mathrm {m^2}+4}{m-2}$

$ =\frac {(m+2)(m-2)-(\ \mathrm {m^2}+4)}{m-2}$

$ =\frac {\ \mathrm {m^2}-4-\ \mathrm {m^2}-4}{m-2}$

$ =-\frac {8}{m-2}$

解：原式$=a-b-(a+b) $

$ =-2b$