一元二次方程7x^2-(m+13)x+m^2-m-2=0两根x1.x2满足0

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一元二次方程7x^2-(m+13)x+m^2-m-2=0两根x1.x2满足0
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一元二次方程7x^2-(m+13)x+m^2-m-2=0两根x1.x2满足0
一元二次方程7x^2-(m+13)x+m^2-m-2=0两根x1.x2满足0

一元二次方程7x^2-(m+13)x+m^2-m-2=0两根x1.x2满足0
楼上 不用考虑判别式吧
令f(x)=7x^2-(m+13)x+m^2-m-2
有f(0)=m^2-m-2>0 且 f(1)= m^2-2m-80
可得 m>2或m

设f(x)=7x^2-(m+13)x+m^2-m-2
满足:0所以:
f(0)=m^2-m-2>0,解得:m>2,or,m<-1
f(1)=7-m-13+m^2-m-2=m^2-2m-8<0,解得:-2f(2)=28-(m+13)*2+m^2-m-2=m^2-3m>0,解得:m>3,or,m<0
判别式=m...

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设f(x)=7x^2-(m+13)x+m^2-m-2
满足:0所以:
f(0)=m^2-m-2>0,解得:m>2,or,m<-1
f(1)=7-m-13+m^2-m-2=m^2-2m-8<0,解得:-2f(2)=28-(m+13)*2+m^2-m-2=m^2-3m>0,解得:m>3,or,m<0
判别式=m^2+26m+169-28(m^2-m-2)>0
-27m^2+54m+225>0
27m^2-54m-225<0
综上所述,3

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∵7x^2-(m+13)x+m^2-m-2=0有两不等实根
∴△=(m+13)^2-4*7*(m^2-m-2)>0 ①
又∵两根x1.x2满足0<x1<1<x2<2且开口向上
∴m^2-m-2>0 ②
7-(m+13)+m^2-m-2<0 ③
28-(m+13)*2+m^2-m-2>0 ④
由①②③④...

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∵7x^2-(m+13)x+m^2-m-2=0有两不等实根
∴△=(m+13)^2-4*7*(m^2-m-2)>0 ①
又∵两根x1.x2满足0<x1<1<x2<2且开口向上
∴m^2-m-2>0 ②
7-(m+13)+m^2-m-2<0 ③
28-(m+13)*2+m^2-m-2>0 ④
由①②③④得:-2<m<-1或3<m<4

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