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题目(2013年全国初中数学联赛试题)已知实数a,b,c,d满足2a2+3c2=2b2+3d2=(ad-bc)2=6,求(a2+b2)(c2+d2)的值.本刊5月下的参考答案同时采用了换元法和夹逼法,意境高,技巧性强.下面提供另两种常规解答,供学习参考.解法1由条件得(2a2+3c2)(2b2+3d2)=36,即4a2b2+6a2 d2+6b2c2+9c2 d2=36.由条件(ad-bc)2=6得a2 d2+b2c2-2abcd=6,∴6a2 d2+6b2c2=36+12abcd.∴4a2b2+12abcd+9c2 d2=0,∴(2ab+3cd)2=0,∴2ab=-3cd.再由条件得2(a2-b2)=-3(c2-d2),上面两式相乘得cd(a2-b2)=ab(c2-d2),
(A2 + b2) (c2 + d2), we know that real numbers a, b, c and d satisfy 2a2 + 3c2 = 2b2 + 3d2 = Value. The publication of May under the reference to the answer at the same time using the method and the law for the surrender, high mood, skillful. The following provides two other conventional solutions for learning reference. Solution 1 from the conditions (2a2 +3 c2) (2b2 + 3d2) = 36, that is, 4a2b2 + 6a2 d2 + 6b2c2 + 9c2 d2 = 36. From the condition (ad-bc) 2 = 6, a2 d2 + b2c2-2abcd = 6 and ∴6a2 d2 + 6b2c2 = 36 + 12abcd (2ab + 3cd) 2 = 0, ∴2ab = -3cd. Then the condition is 2 (a2-b2) = -3 (c2-d2), the above two phases Multiply cd (a2-b2) = ab (c2-d2),