!! +var+ +exec+ +string+
$lst_tit solve$lst_titend$type_both
  $lst_alias rootof
  $(lst_syntax)solve $emph fn$emphend for $emph v$emphend=$emph v1$emphend
  to $emph v2$emphend</tt>
  $lst_mean 找出变量 $emph v$emphend 在范围 $emph v1$emphend 与
  $emph v2$emphend 内, 函数或方程 $emph fn$emphend 的(实单)根. <p>
  $emph fn$emphend 在区间 [v1,v2] 内可能有几个根, 但太接近的根可能被忽视.

