已知a²+b²=2,c²+d²=2,ac=bd,求证:a²+c²=2,b²+d²=2,ab=cd
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![已知a²+b²=2,c²+d²=2,ac=bd,求证:a²+c²=2,b²+d²=2,ab=cd](/uploads/image/z/4078540-28-0.jpg?t=%E5%B7%B2%E7%9F%A5a%26%23178%3B%2Bb%26%23178%3B%3D2%2Cc%26%23178%3B%2Bd%26%23178%3B%3D2%2Cac%3Dbd%2C%E6%B1%82%E8%AF%81%EF%BC%9Aa%26%23178%3B%2Bc%26%23178%3B%3D2%2Cb%26%23178%3B%2Bd%26%23178%3B%3D2%2Cab%3Dcd)
已知a²+b²=2,c²+d²=2,ac=bd,求证:a²+c²=2,b²+d²=2,ab=cd
已知a²+b²=2,c²+d²=2,ac=bd,求证:a²+c²=2,b²+d²=2,ab=cd
已知a²+b²=2,c²+d²=2,ac=bd,求证:a²+c²=2,b²+d²=2,ab=cd
设a=bt,d=ct,则b^2(1+t^2)=2,c^2(1+t^2)=2,得b^2=c^2