已知不等式f(x)=x2+px+q

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已知不等式f(x)=x2+px+q
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已知不等式f(x)=x2+px+q
已知不等式f(x)=x2+px+q

已知不等式f(x)=x2+px+q
x2+px+q

2,5是f(x)=x2+px+q=0的解.
p=-(2+5)=-7
q=2*5=10
4x/f(x)≤1
4x/(x^2-7x+10)<=1
[(x^2-7x+10)-4x]/(x^2-7x+10)>=0
(x^2-11x+10)/(x^2-7x+10)>=0
(x^2-11x+10)*(x^2-7x+10)>=0
(x-1)(x-10)(x-2)(x-5)>=0
x<=1或2<=x<=5,或x>=10

4x/[(x-2)(x-5)]<=1
因为f(x)=x2+px+q<0
x^2-7x+10<=4x
x^2-11x+10<=0
1<=x<=10
又因为f(x)=x2+px+q<0的解集为(2,5)
所以2