初二数学吧 关注：14,669贴子：68,627
- 4回复贴，共1页
提示 由0 = 2(ab + cd -6) + 2(ac + bd -3) + (a + b + c + d)(a + b + c + d - 6) = (a + d - 3)² + (b + c - 3)², 得降次式a + d = b + c = 3. 再用4abcd = (ad + bc - 2)(ac + bd + a + d + 3) - (ab + cd - 6)(ac - ad - bc + bd - a - d - 1) + (a + d - 3)(abc - abd - acd + bcd - ab - ad - bc - cd - b - c - 1) - (b + c - 3)(abc - abd - acd + bcd + 3a + 3d + 3) = 0减元求解.
