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由 galaxylee 於 星期五 十二月 29, 2006 1:25 am
#ed_op#DIV#ed_cl##ed_op#DIV class=bd#ed_cl##ed_op#DIV class="main breakfix"#ed_cl##ed_op#P#ed_cl#(b^2+c^2-a^2/2bc)+(a^2+c^2-b^2/2ac)+(b^2+a^2-c^2/2ab)=1#ed_op#BR#ed_cl#=> a(b^2+c^2-a^2)+b(a^2+c^2-b^2)+c(b^2+a^2-c^2)=2abc#ed_op#BR#ed_cl#=> ab^2+ac^2-a^3+ba^2+bc^2-b^3+cb^2+ca^2-c^3-2abc=0#ed_op#BR#ed_cl#=> (ba^2+ca^2-a^3)+(ab^2-2abc+ac^2)+(bc^2-c^3+cb^2-b^3)=0#ed_op#BR#ed_cl#=> a^2(b+c-a)+a(b-c)^2-(b+c)(b-c)^2=0#ed_op#BR#ed_cl#=> a^2(b+c-a)-(b+c-a)(b-c)^2=0#ed_op#BR#ed_cl#=> (b+c-a)(a+b-c)(a-b+c)=0#ed_op#BR#ed_cl#=> a=b+c 或 a=c-b 或 a=b-c#ed_op#/P#ed_cl##ed_op#P#ed_cl#a用b+c或c-b或b-c代回去即可得證#ed_op#/P#ed_cl##ed_op#P#ed_cl# #ed_op#/P#ed_cl##ed_op#/DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#/DIV#ed_cl#
[數學]數學題..
由 skywalker 於 星期四 十二月 28, 2006 11:13 pm
#ed_op#DIV#ed_cl#(a#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#+b#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#-c#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#)/2ab+(a#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#+c#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#-b#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#)/2ac+(b#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#+c#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#-a#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#)/2bc=1#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#求證其中兩分式等於1，另一個等於-1#ed_op#/DIV#ed_cl#