Cho đa thức: ax³+bx² +cx+d (a;b;c;d ∈ R) Biết 13a-6b+4c+0. Chứng minh P(1/2).P(-2) ≥ 0

Giải thích các bước giải:

Ta có:

$P(x)=ax^3+bx^2+cx+d$

$\to \begin{cases} P(\dfrac12)=a\cdot (\dfrac12)^3+b\cdot (\dfrac12)^2+c\cdot \dfrac12+d\\ P(-2)=a\cdot (-2)^3+b\cdot (-2)^2+c\cdot (-2)+d\end{cases}$

$\to \begin{cases} P(\dfrac12)=\dfrac18a+\dfrac14b+\dfrac12c+d\\ P(-2)=-8a+4b-2c+d\end{cases}$

$\to \begin{cases} 8P(\dfrac12)=a+2b+4c+8d\\ -P(-2)=8a-4b+2c-d\end{cases}$

$\to \begin{cases} 8P(\dfrac12)=a+2b+4c+8d\\ -2P(-2)=16a-8b+4c-2d\end{cases}$

Mà $13a-6b+4c=0\to 4c=-13a+6b$

$\to \begin{cases} 8P(\dfrac12)=a+2b+(-13a+6b)+8d\\ -2P(-2)=16a-8b+(-13a+6b)-2d\end{cases}$

$\to \begin{cases} 8P(\dfrac12)=-12a+8b+8d\\ -2P(-2)=3a-2b-2d\end{cases}$

$\to \begin{cases} 2P(\dfrac12)=-3a+2b+2d\\ -2P(-2)=3a-2b-2d\end{cases}$

$\to \begin{cases} 2P(\dfrac12)=-3a+2b+2d\\ 2P(-2)=-3a+2b+2d\end{cases}$

$\to 2P(\dfrac12)\cdot 2P(-2)=(-3a+2b+2d)^2\ge 0$

$\to 4P(\dfrac12)\cdot P(-2)\ge 0$

$\to P(\dfrac12)\cdot P(-2)\ge 0$

$\to đpcm$