已知二次函数y=-x^2+X-1/5,当自变量x取m时,对应的函数值大于0,当自变量x分别取m-1 m+1时对应的函数值值y1,y2满足?A、y1>0,y2>0 B、y1<0,y2<0C、y1<0,y2>0 C、y1>0,y2<0
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![已知二次函数y=-x^2+X-1/5,当自变量x取m时,对应的函数值大于0,当自变量x分别取m-1 m+1时对应的函数值值y1,y2满足?A、y1>0,y2>0 B、y1<0,y2<0C、y1<0,y2>0 C、y1>0,y2<0](/uploads/image/z/2498968-64-8.jpg?t=%E5%B7%B2%E7%9F%A5%E4%BA%8C%E6%AC%A1%E5%87%BD%E6%95%B0y%3D-x%5E2%2BX-1%2F5%2C%E5%BD%93%E8%87%AA%E5%8F%98%E9%87%8Fx%E5%8F%96m%E6%97%B6%2C%E5%AF%B9%E5%BA%94%E7%9A%84%E5%87%BD%E6%95%B0%E5%80%BC%E5%A4%A7%E4%BA%8E0%2C%E5%BD%93%E8%87%AA%E5%8F%98%E9%87%8Fx%E5%88%86%E5%88%AB%E5%8F%96m-1+m%2B1%E6%97%B6%E5%AF%B9%E5%BA%94%E7%9A%84%E5%87%BD%E6%95%B0%E5%80%BC%E5%80%BCy1%2Cy2%E6%BB%A1%E8%B6%B3%3FA%E3%80%81y1%EF%BC%9E0%2Cy2%EF%BC%9E0+B%E3%80%81y1%EF%BC%9C0%2Cy2%EF%BC%9C0C%E3%80%81y1%EF%BC%9C0%2Cy2%EF%BC%9E0+C%E3%80%81y1%EF%BC%9E0%2Cy2%EF%BC%9C0)
已知二次函数y=-x^2+X-1/5,当自变量x取m时,对应的函数值大于0,当自变量x分别取m-1 m+1时对应的函数值值y1,y2满足?A、y1>0,y2>0 B、y1<0,y2<0C、y1<0,y2>0 C、y1>0,y2<0
已知二次函数y=-x^2+X-1/5,当自变量x取m时,对应的函数值大于0,当自变量x分别取m-1 m+1时对应的函数值
值y1,y2满足?
A、y1>0,y2>0 B、y1<0,y2<0
C、y1<0,y2>0 C、y1>0,y2<0
已知二次函数y=-x^2+X-1/5,当自变量x取m时,对应的函数值大于0,当自变量x分别取m-1 m+1时对应的函数值值y1,y2满足?A、y1>0,y2>0 B、y1<0,y2<0C、y1<0,y2>0 C、y1>0,y2<0
把m-1 m+1分别代入y=-x^2+X-1/5,得到-m^2+m-1/5-2①,-m^2-m-1/5②,又因为-m^2+m-1/5>0,所以①>0,②
令 y=0,
解得:x=(5+根号5)/10 ,x=(5-根号5)/10
∵当自变量x取m时对应的值大于0,
∴ (5-根号5)/10<m<(5+根号5)/10 ,
∴m-1<(5-根号5)/10 ,m+1>(5+根号5)/10 ,
∴y1<0、y2<0.
故选B.
(【(5+根号5)/10】-【(5-根号5)/10】小于1 )...
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令 y=0,
解得:x=(5+根号5)/10 ,x=(5-根号5)/10
∵当自变量x取m时对应的值大于0,
∴ (5-根号5)/10<m<(5+根号5)/10 ,
∴m-1<(5-根号5)/10 ,m+1>(5+根号5)/10 ,
∴y1<0、y2<0.
故选B.
(【(5+根号5)/10】-【(5-根号5)/10】小于1 )
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