高中二次函数二次函数f(x)=p
m>0，
∴m/(m+2)-[m/(m+1)]^2=m/[(m+2)(m+1)^2]>0。
由p/(m+2)+q/(m+1)+r/m=0得
r=-m[p/(m+2)+q/(m+1)],
∴f(x)=p[x^2-m/(m+2)]+q[x-m/(m+1)],
f[m/(m+1)]=p{[m/(m+1)]^2-m/(m+2)},
f{√[m/(m+2)]}=q{√[m/(m+2)]-m/(m+1)},
当pq>=0时，f[m/(m+1)]*f{√[m/(m+2)]}-q(m+2)/(m+1)或p0,f(1)>0,p>0,
则f[m/(m+1)]p>-q(m+2)/(m+1)时f(0)>0,f(1...全部
m>0，
∴m/(m+2)-[m/(m+1)]^2=m/[(m+2)(m+1)^2]>0。
由p/(m+2)+q/(m+1)+r/m=0得
r=-m[p/(m+2)+q/(m+1)],
∴f(x)=p[x^2-m/(m+2)]+q[x-m/(m+1)],
f[m/(m+1)]=p{[m/(m+1)]^2-m/(m+2)},
f{√[m/(m+2)]}=q{√[m/(m+2)]-m/(m+1)},
当pq>=0时，f[m/(m+1)]*f{√[m/(m+2)]}-q(m+2)/(m+1)或p0,f(1)>0,p>0,
则f[m/(m+1)]p>-q(m+2)/(m+1)时f(0)>0,f(1)>0,p0,方程f(x)=0在(0,1)内无解。
命题不成立。
。收起
