$?解：(1)原式 ={\frac {(1+2x)(1-2x)} {x(2x+1)}}$
$ ={\frac {1-2x} {x}}$
$將x=-1代入，得$
$原式 ={\frac {1-2×(-1)} {-1}}$
$ =-3$
$(2) 原式 ={\frac {y({x}^{2}-2x+1)} {y(1-{x}^{2})}}$
$ ={\frac {y{(1-x)}^{2}} {y(1+x)(1-x)}}$
$ ={\frac {1-x} {1+x}}$
$將x=2代入，得$
$原式 ={\frac {1-2} {1+2}}$
$ =-{\frac {1} {3}}$
$(3) 原式 ={\frac {a(a-3b)} {{(a-3b)}^{2}}}$
$ ={\frac {a} {a-3b}}$
$將a=\frac 3 2、b=-\frac 2 3代入，得$
$原式 ={\frac {\frac {3} {2}} {\frac {3} {2}-3×(-{\frac {2} {3}})}}$
$ ={\frac {3} {7}}$
