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Xét $f(x)=ax^3+bx+c$.
Ta có: $f(1)=a+b+c$
$f(\frac{1}{2})=\frac{a}{8}+\frac{b}{2}+c$
$f(0)=c$
Suy ra: $f(1)+4f(\frac{1}{2})+f(0)=\frac{3}{2}(a+2b+4c)=0$
*) Nếu $f(\frac{1}{2})=0$ , đpcm.
*) Nếu $f(\frac{1}{2})\ne0 \Rightarrow \left[ \begin{array}{l} f(1)f(\frac{1}{2})<0\\ f(0)f(\frac{1}{2})<0 \end{array} \right.$, đpcm.
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