已知a²+b²=5,c²+d²=2,求(ac+bd)²+(ad-bc)²的值
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![已知a²+b²=5,c²+d²=2,求(ac+bd)²+(ad-bc)²的值](/uploads/image/z/7612840-64-0.jpg?t=%E5%B7%B2%E7%9F%A5a%26%23178%3B%2Bb%26%23178%3B%3D5%2Cc%26%23178%3B%2Bd%26%23178%3B%3D2%2C%E6%B1%82%EF%BC%88ac%2Bbd%EF%BC%89%26%23178%3B%2B%EF%BC%88ad-bc%EF%BC%89%26%23178%3B%E7%9A%84%E5%80%BC)
已知a²+b²=5,c²+d²=2,求(ac+bd)²+(ad-bc)²的值
已知a²+b²=5,c²+d²=2,求(ac+bd)²+(ad-bc)²的值
已知a²+b²=5,c²+d²=2,求(ac+bd)²+(ad-bc)²的值
原式=a²c²+2abcd+b²c²+a²d²-2abcd+b²c²
=(a²c²+a²d²)+(b²c²+b²c²)
=a²(c²+d²)+b²(c²+d²)
=(c²+d²)(a²+c²)
=2×5
=10
由题得:
=a²c²+2abcd+b²c²+a²d²-2abcd+b²c²
=(b²c²+b²c²)+(a²c²+a²d²)
=b²(c²+d²)+a²(c²+d²)
=(a²+c²)(c²+d²)
=2×5
=10
(ac+bd)²+(ad-bc)²
=(ac)^2+(bd)^2+(ad)^2+(bc)^2
=a^2(c^2+d^2)+b^2(c^2+d^2)
=(a^2+b^2)(c^2+d^2)
=10
(ac+bd)²+(ad-bc)²=(a²+b²)*(c²+d²)=5*2=10
答案为10 ,要过程不?