已知 4a－3b－6c = 0，且 a＋2b－7c = 0，求代數式 (2a2＋3b2＋6c2)/(a2＋5b2＋7c2) 之值。
2005-11-22 15:53:32 · 1 個解答 · 發問者 Frank 7 in 科學 ➔ 數學
4a－3b－6c = 0......(1) a＋2b－7c = 0.......(2)(2)*4-(1)11b-22c=0.b=2c代入(2)a+4c-7c=0.a=3c.a:b:c:=3:2:1(2a2＋3b2＋6c2)/(a2＋5b2＋7c2) =(18+12+6)/(9+20+7)=36/36=1.為解!
2005-11-22 16:03:49 · answer #1 · answered by Anonymous · 0⤊ 0⤋