二次函数f(x)=px^2+qx+r中实数p、q、r满足p/(m+2)+q/(m+1)+r/m=0,其中m>0.求证：（1）pf(m/(m+1))扫码下载作业帮拍照答疑一拍即得

∵p/(m+2)+q/(m+1)+r/m=0
∴p^2/(m+2)+pq/(m+1)+pr/m=0
∴p^2m/(m+2)+pqm/(m+1)+pr=0
∵pf(m/(m+1))=p^2(m/(m+1))^2+pq(m/(m+1))+pr
=p^2(m/(m+1))^2-p^2m/(m+2)
=-p^2m/((m+1)^2(m+1))
∵p^2m＞0,(m+1)^2＞0,又m＞0
∴-m/(m+1)＜0
∴pf(m/(m+1))

p/(m+2)+q/(m+1)+r/m=0
ppm/(m+2)+mpq/(m+1)+pr=0
pf(m/(m+1))
＝ppmm/(m+1)^2+pqm/(m+1)+pr
＝ppmm/(m+1)^2－ppm/(m+2)
＝ppmm/(m+1)^2－ppmm/(m+2)m
＝ppmm[1/(m+1)^2-1/(m+2)m]
(m+1)^2>(m+2)m>0
1/(m+1)^2-1/(m+2)mmm>0
二次函数f(x)=px^2+qx+r
p不＝0
pp>0
ppmm[1/(m+1)^2-1/(m+2)m]

∵p/(m+2)+q/(m+1)+r/m=0
∴(p^2)m/(m+2)+pqm/(m+1)+pr=0
∴pqm/(m+1)+pr=-(p^2)m/(m+2)
pf(m/(m+1))
=p[p[m/(m+1)]^2+q[m/(m+1)]+r]
=[pm/(m+1)]^2+pqm/(m+1)+pr
=[pm/(m+1)]^2-(p^2)m/(m+2) {代入上式结论}
=(p^2)m[m/(m+1)^2-1/(m+2)]
=(p^2)m{[m(m+2)-(m+1)^2]/(m+2)(m+1)^2}
=(p^2)m[(-1)/(m+2)(m+1)^2]
=-(p^2)m[1/(m+2)(m+1)^2]