Question

请教一道数学题～～关于乘法公式的!

已知:a平方+b平方=c平方+d平方=1,
求证:(ac-bd)的平方+(ad+bc)的平方=1.
帮忙解一下,最好解释一下为什么?

Answer 1

a^2+b^2=c^2+d^2=1
(a^2+b^2)(c^2+d^2)=1
(ac)^2+(bc)^2+(ad)^2+(bd)^2=1
(ac)^2+(bd)^2+(ad)^2+(bc)^2=1
(ac)^2-2acbd+(bd)^2+(ad)^2+2adbc+(bc)^2=1
(ac-bd)^2+(ad+bc)^2=1

Answer 2

(ac-bd)^2+(ad+bc)^2
=ac^2+bd^2-2abcd+ad^2+bc^2+2abcd
=ac^2+bd^2+bc^2+ad^2
=(a^2+b^2)(c^2+d^2)=1*1=1

Answer 3

(ac-bd)^2+(ad+bc)^2=ac^2-2*abcd+bd^2+ad^2+2abcd+bc^2
=ac^2+bd^2+ad^2+bc^2
=(a^2+b^2)(c^2+d^2)
=1*1
=1