Fie \( f:\mathbb{R}\rightarrow \mathbb{R},\ f(x)=x^3+ax^2+bx+c \) si \( m,n,p \) solutiile ecuatiei \( f(x)=0 \), reale si distincte. Sa se demonstreze ca \( \frac{m^2}{f^{\prime}(m)}+\frac{n^2}{f^{\prime}(n)}+\frac{p^2}{f^{\prime}(p)}=1 \).
Mie unul mi s-a parut grea tinand cont ca nu am vazut alta metoda decat cea "muncitoreasca"
Identitate cu solutiile unei ecuatii de gradul 3
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Laurentiu Tucaa
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mihai miculita
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\( f(x)=x^3+ax^2+bx+c=(x-m)(x-n)(x-p)\Rightarrow f^{\prime}(x)=(x-n)(x-p)+(x-m)(x-p)+(x-m)(x-n)\Rightarrow \left\{\begin f^{\prime}
(m)=(m-n)(m-p)\\
f^{\prime}(n)=(n-m)(n-p)\\
f^{\prime}(p)=(p-m)(p-n)\right \)
\( \Rightarrow \frac{m^2}{f^{\prime}(m)}+\frac{n^2}{f^{\prime}(n)}+\frac{p^2}{f^{\prime}(p)}=\frac{m^2}{(m-n)(m-p)}+\frac{n^2}{(n-m)(n-p)}+\frac{p^2}{(p-m)(p-n)}=\frac{(p-n)m^2+(m-p)n^2+(n-m)p^2}{(m-n)(n-p)(p-m)}=\frac{(m-n)(n-p)(p-m)}{(m-n)(n-p)(p-m)}=1; \)
\( \mbox{fiindca: }\\
(p-n)m^2+(m-p)n^2+(n-m)p^2=p(m^2-n^2)-mn(m-n)-(m-n)p^2=p(m-n)(m+n)-mn(m-n)-(m-n)p^2=\\
=(m-n).[-p^2+(m+n)p-mn]=(m-n)[(n-p)p-m(n-p)]=(m-n)(n-p)(p-m). \)
Intr-un mesaj privat mi s-a cerut sa fac aceasta ultima detaliere...; desi verificarea penultimei egalitati era mai usoara pornind de la produsul: (m-n)(n-p)(p-m)=...
(m)=(m-n)(m-p)\\
f^{\prime}(n)=(n-m)(n-p)\\
f^{\prime}(p)=(p-m)(p-n)\right \)
\( \Rightarrow \frac{m^2}{f^{\prime}(m)}+\frac{n^2}{f^{\prime}(n)}+\frac{p^2}{f^{\prime}(p)}=\frac{m^2}{(m-n)(m-p)}+\frac{n^2}{(n-m)(n-p)}+\frac{p^2}{(p-m)(p-n)}=\frac{(p-n)m^2+(m-p)n^2+(n-m)p^2}{(m-n)(n-p)(p-m)}=\frac{(m-n)(n-p)(p-m)}{(m-n)(n-p)(p-m)}=1; \)
\( \mbox{fiindca: }\\
(p-n)m^2+(m-p)n^2+(n-m)p^2=p(m^2-n^2)-mn(m-n)-(m-n)p^2=p(m-n)(m+n)-mn(m-n)-(m-n)p^2=\\
=(m-n).[-p^2+(m+n)p-mn]=(m-n)[(n-p)p-m(n-p)]=(m-n)(n-p)(p-m). \)
Intr-un mesaj privat mi s-a cerut sa fac aceasta ultima detaliere...; desi verificarea penultimei egalitati era mai usoara pornind de la produsul: (m-n)(n-p)(p-m)=...