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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.

题目详情
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.
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答案和解析
∵2a2+3c2=2b2+3d2=6,
∴(2a2+3c2)(2b2+3d2)=36,即4a2b2+6a2d2+6b2c2+9c2d2=36,
∵(ad-bc)2=6,
∴a2d2-2abcd+b2c2=6,即a2d2+b2c2=6+2abcd,
∴6a2d2+6b2c2=36+12abcd,
∴4a2b2+36+12abcd+9c2d2=36,即(2ab+3cd)2=0,
∴2ab=-3cd ①,
由2a2+3c2=2b2+3d2知2(a2-b2)=-3(c2-d2) ②,
①×②,得:-6cd(a2-b2)=-6ab(c2-d2),即cd(a2-b2)=ab(c2-d2),
a2cd-b2cd-abc2+abd2=0,
(ac+bd)(ad-bc)=0,
∵(ad-bc)2=6,
∴ad-bc≠0,
则ac+bd=0,
∴(a2+b2)(c2+d2)=a2c2+b2d2+a2d2+b2c2
=(ac+bd)2+(ad-bc)2
=02+6
=6.