初一假期数学作业,急!帮帮忙!

（1）在三角形ABC中,角ABC的平分线和角ACB的外角平分线相交于点D,若角D=40°,则角BAC的大小为（ D ）.         A.50°  B.60°  C.70°  D.80°理由:先设∠ABD为X,那么∠DBC也为X了(理由略)     因为DC平分∠ACB的外角     所以∠ACD=0.5∠ACB的外角              =0.5(∠A+2X)(三角形外角等于与他不相临的两内角和)               =0.5∠A+X     因为∠D=180°-X-∠ACB-0.5∠A-X            =180°-2X-∠ACB-0.5∠A    又因为∠ABC+∠ACB+∠A=180°     所以180°-2X-∠ACB-0.5∠A=40°         180°-∠ABC-∠ACB-0.5∠A=40°              180°- 180°+0.5∠A=40°                           0.5∠A=40°                              ∠A=80°（2）如图：A、B、C、D、F据为有理数,图中各行、各列及两条对角线上三个数之和相等,试求（AB+CD+EF）÷（A+B+C+D+E+F）的值.      如我画的图可知:A+B+6=C+D+E=F+7+2=A+D+2=6+D+F=A+C+F=B+D+7=6+1+2                    因为F+7+2=6+D+F                    所以F+9=6+D+F                          9=6+D                          D=3                    因为D=2所以A+D+2=A+3+2=A+B+6                                         B=-1                    因为D=3,B=-1                    所以B+D+7=9                    所以A+D+2=A+3+2=9,A=4                        F+7+2=9,F=0                        6+E+2=9,E=1                        C+D+E=C+3+1=9,C=5                      （AB+CD+EF）÷（A+B+C+D+E+F）=11/12已知a+3b+c=0(1)  3a-3b-4c=0(2),求a:b:c.(1)+(2)=4A-3C=0           4A=3C           C=4/3A将C=4/3A代入(2)得          B=-7/9AA:B;C=9:12:-74)一列快车长168米,一列慢车长184米,如果两车相向而行,从相遇到离开需4秒,如果同向而行,从快车追上慢车到离开需16秒,求两车速度.           解设快车速度XM/秒,慢车速度YM/秒.            根据题意得:4(X+Y)=168+184                       16(X-Y)=168+184                   解得X=55,Y=33