是不是求ac-bd啊?
(ad+bc)²+(ac-bd)²
=a²d²+2abcd+b²c²+a²c²-2abcd+b²d²
=a²d²+b²c²+a²c²+b²d²
=(a²+b²)(c²+d²)
=1*1
=1
所以
(ac-bd)²=1-(ad+bc)²=1
ac-bd=1或ac-bd=-1
是不是求ac-bd啊?
(ad+bc)²+(ac-bd)²
=a²d²+2abcd+b²c²+a²c²-2abcd+b²d²
=a²d²+b²c²+a²c²+b²d²
=(a²+b²)(c²+d²)
=1*1
=1
所以
(ac-bd)²=1-(ad+bc)²=1
ac-bd=1或ac-bd=-1