请问有没有办法，对日期的年月直接进行加减操作？能不能提供一个小例子？(20分)

直接？？用转化吧：strtodate()

有啊，D6中有，我将D5卸载了。
IncAMonth procedure

Increments date data by one month.

IncDay function

Returns a date shifted by a specified number of days.

IncHour function

Returns a date/time value shifted by a specified number of hours.

IncMilliSecond function

Returns a date/time value shifted by a specified number of milliseconds.

IncMinute function

Returns a date/time value shifted by a specified number of minutes.

IncMonth function

Returns a date shifted by a specified number of months.

IncSecond function

Returns a date/time value shifted by a specified number of seconds.

IncWeek function

Returns a date shifted by a specified number of weeks.

IncYear function

Returns a date shifted by a specified number of years.

在Delphi中，tdatetime本身就是基于tfloat的，直接加减就行了
example1:
T:TDateTime;
T:=T+StrToTime('00:00:01');//T加一秒。
T:=T+StrToTime('00:01:00');//T加一分钟。
在SQL中，用dateadd函数
DATEADD(day, 21, pubdate)返回pubdate（datetime）加21天，具体你可以看看query analyzer
中t-sql的帮助

TDATTIME在delphi中其实就是一个浮点类型,看一看帮助！

两个函数 function EncodeDate(year,month,day:word);
可把三个word型变量转换成一个日期型时间变量;
procedure DecodeDate(date:tdatetime;var year,month,day:word);
可把日期型(####年##月##日)变为三个word型变量.
var date1,date2,date:tdatetime;
year1,year2,month1,month2,day1,day2,year,month,day:word;
程序部分
date1:=encode(year1,month1,day1);
date2:=encode(year2,month2,day2);
date:=date1-date2;
decodedate(date,year,month,day);
edit1.text:='两个日期之间间隔'+inttostr(year-1900)+'年'=inttostr(month)+'月'+inttostr(day)+'日';

begindate,enddate:variant;
date_1,date_2:Tdatetime;
days:integer;

begindate:=date_1;
enddate:=date_2;
days:=enddate-begindate;

days就是两个日期的间隔天数了。。：）

TDATTIME在delphi中是一个浮点类型
var
t1,t2:Tdatetime;
i:real;
begin
t1:=now;
t2:=now+1; //加一天
i:=t2-t1; //两个日期型相减，相差为 1天

D6的函数：
procedure IncAMonth(var Year, Month, Day: Word; NumberOfMonths: Integer = 1);

Description

IncAMonth modifies Year, Month, and Day parameters so the date they describe
is incremented by NumberOfMonths months. NumberOfMonths can be negative,
to return a date N months previous.

If the input day of month is greater than the last day of the resulting month,
the day is set to the last day of the resulting month.