如图,抛物线y=x^2+bx+c与x轴交于A、B两点,与y轴交于点C,∠OBC=45°,则下列各式成立的是 A b-c-1=0 B b+c-1=0 C b-c+1=0 D b+c+1=0 答案选D,不知道为什么
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![如图,抛物线y=x^2+bx+c与x轴交于A、B两点,与y轴交于点C,∠OBC=45°,则下列各式成立的是 A b-c-1=0 B b+c-1=0 C b-c+1=0 D b+c+1=0 答案选D,不知道为什么](/uploads/image/z/3816129-57-9.jpg?t=%E5%A6%82%E5%9B%BE%2C%E6%8A%9B%E7%89%A9%E7%BA%BFy%3Dx%5E2%2Bbx%2Bc%E4%B8%8Ex%E8%BD%B4%E4%BA%A4%E4%BA%8EA%E3%80%81B%E4%B8%A4%E7%82%B9%2C%E4%B8%8Ey%E8%BD%B4%E4%BA%A4%E4%BA%8E%E7%82%B9C%2C%E2%88%A0OBC%3D45%C2%B0%2C%E5%88%99%E4%B8%8B%E5%88%97%E5%90%84%E5%BC%8F%E6%88%90%E7%AB%8B%E7%9A%84%E6%98%AF+++A+++b-c-1%3D0++++B++b%2Bc-1%3D0++++C++b-c%2B1%3D0+++D+b%2Bc%2B1%3D0+++++++%E7%AD%94%E6%A1%88%E9%80%89D%2C%E4%B8%8D%E7%9F%A5%E9%81%93%E4%B8%BA%E4%BB%80%E4%B9%88)
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