由 路人 於 星期二 七月 24, 2007 7:52 am
#ed_op#DIV#ed_cl#(x+y+1)/[x(y+1)]=1/2#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#1/x+1/(y+1)=1/2.........(1)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#(x+z+2)/[x(z+2)]=1/3#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#1/x+1/(z+2)=1/3.........(2)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#[(y+1)+(z+2)]/[(y+1)(z+2)]=1/4#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#1/(y+1)+1/(z+2)=1/4...(3)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#(1)+(2)+(3)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#1/x+1/(y+1)+1/(z+2)=13/24.......(4)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#(4)-(1)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#1/(z+2)=1/24#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#z=22#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#