设a,b,c,d是实数,且ad-bc=1,a²+b²+c²+d²-ab+cd=1,求abcd的值

kuai
2025-07-01 07:14:17
推荐回答(1个)
回答1:

2a²+2b²+2c²+2d²-2ab+2cd=2
2a²+2b²+2c²+2d²-2ab+2cd=2(ad-bc)

2a²+2b²+2c²+2d²-2ab+2cd-2ad+2bc=0
(a²-2ab+b²)+(c²+2cd+d²)+...=0
(a-b)²+(c+d)²+(a-d)²+(b+c)²=0
得出a=b=d=-c
ad-bc=1即 2a²=1
a²=0.5
abcd=-aaaa=-0.25