方法一：当x＝17（或18或19）时，y1＋y2＜0． ··················································· 5分 方法二：当y1＋y2＝0时，x＝55＋5（负根舍去）．
又因为当55＋5＜x＜20时，y1＋y2随x的增大而减小，
所以当x＝17或18或19时，y1＋y2＜0． ·················································· 5分
（2）设第一个月单价降低x元时，销售完这批T恤获得的利润为1000元．
根据题意得－10x2＋100x＋1000＝1000． ································································ 6分 解这个方程，得x1＝0，x2＝10．
从增加销售量的角度看，取x＝10． ······································································ 7分 答：第一个月单价降低10元时，销售完这批T恤获得的利润为1000元． ····················· 8分
26．（本题10分）
3π
－． ····································································· 8分 26
10
解：（1）方法一：
如图，设半圆O2与BD 的切点为E，连接O2E，则O2E⊥BD． ······························· 1分 OA D 1 ∵半圆O2与CD 相切，且∠C＝90°， ∴O2E＝O2C，DC＝DE＝a．
在Rt△BEO2中，O2B2＝BE2＋O2E2． ·················· 2分 ∴(b－EO2)2＝(a2＋b2－a)2＋O2E2． ·················· 3分
aa2＋b2－a2aa2＋b2－a2 解得EO2＝，所以最终拼接成的圆形桌面的半径为bb
m． ············································································································ 4分 方法二：
如图，设半圆O2与BD 的切点为E，连接O2E，则O2E⊥BD． ······························· 1分 ∵半圆O2与CD 相切，且∠C＝90°， ∴O2E＝O2C．
∵∠EBO2＝∠CBD，∠BEO2＝∠BCD＝90°，
∴△BEO2∽△BCD． ······················································································· 2分 EO2BO2
∴＝．
CDBD
EO2b－EO2
∴＝22． ·························································································· 3分
aa＋b
aa2＋b2－a2aa2＋b2－a2 解得EO2＝，所以最终拼接成的圆形桌面的半径为bb
m． ····················································································································· 4分
（说明：求出最终拼接成的圆形桌面的半径为
ab
m不扣分．）
a2＋b2＋a
B O2
C E （2）①小明的错误是半圆O1与半圆O2不能保证外切，即“O1O2＝2x”是错误的．
····································································································································· 7分
②方法一：
要使小明解得的答案是正确的，就要半圆O1与半圆O2外切．
此时半圆O1与BD 的切点、半圆O2与BD的切点以及O1O2与BD的交点重合．
所以a2＋b2－a＝a． ··················································································· 8分
解得b＝3a． ·························································································· 10分 方法二：
aa2＋b2－a2a2＋b2令＝． ············································································· 8分
b4b4aa2＋b2－4a2＝a2＋b2． a2＋b2－4aa2＋b2＋4a2＝0．
11
即(a2＋b2－2a) 2＝0．
解得b＝3a． ·························································································· 10分
27．（本题12分）
解：方案预设
（1）预设方案1：
①15． ········································································································· 2分
②方法一：
设王老师把乙送到车站后，再经过m h与甲相遇．
1
(45＋5)m＝15－5×． ·················································································· 3分