Chứng minh rằng ít nhất một trong ba phương trình sau có nghiệm:
$(m-1)x^{2}+2(n+1)x+p=0    (1)$
$(n+1)x^{2}+2px+(m-1)=0   (2)$
$px^{2}+2(m-1)x+(n+1)=0   (3)$
Ta có $\Delta^{'}_{1}=(n+1)^{2}-(m-1)p=n^{2}+2n+1-mp+p$
$\Delta^{'}_{2}=p^{2}-(n+1)(m-1)=p^{2}-mn-m+n+1$
$\Delta^{'}_{3}=(m-1)^{2}-p(n+1)=m^{2}-2m+1-pn-p$
$\Delta^{'}_{1}+\Delta^{'}_{2}+\Delta^{'}_{3}=\frac{1}{2}(m^{2}+n^{2}+4-2mn-4m+4n)+$
$\frac{1}{2}(n^{2}+p^{2}+1-2n-2np-2p)+\frac{1}{2}(p^{2}+m^{2}+1-2pm-2p-2m)$
$=\frac{1}{2}[(m-n-2)^{2}+(n-p+1)^{2}+(p-m+1)^{2}]>0$
Vậy trong đó phải có một $\Delta^{'}>0 $ nên một trong ba  phương trình đã cho có ít nhất một phương trình có nghiệm

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