Question

cho đa thức P(x) = \(ax^3+bx^2+cx+d\) . Biết 13a-6b+4c=0 . CMR : \(P\left(\dfrac{1}{2}\right).P\left(-2\right)\ge0\)

Answer 1

Lời giải:

Ta có:

\(P\left(\frac{1}{2}\right)=\frac{a}{8}+\frac{b}{4}+\frac{c}{2}+d=\frac{1}{8}(a+2b+4c+8d)\)

\(\Rightarrow 8P\left(\frac{1}{2}\right)=a+2b+4c+8d(1)\)

\(P(-2)=-8a+4b-2c+d\)

\(\Rightarrow 8P(-2)=-64a+32b-16c+8d(2)\)

Từ \((1); (2)\Rightarrow 8P(\frac{1}{2})-8P(-2)=(a+2b+4c+8d)-(-64a+32b-16c+8d)\)

\(=65a-30b+20c\)

\(=5(13a-6b+4c)=0\)

Do đó: \(8P(\frac{1}{2})=8P(-2)\Leftrightarrow P(\frac{1}{2})=P(-2)\)

\(\Rightarrow P(\frac{1}{2})P(-2)=[P(-2)]^2\geq 0\)

Ta có đpcm.