\(\left\{{}\begin{matrix}a^2\ge0\\a^4+a^2+1>0\end{matrix}\right.\) ;\(\forall a\Rightarrow P=\dfrac{a^2}{a^4+a^2+1}\ge0\)
\(P_{min}=0\) khi \(a=0\)
\(P=\dfrac{3a^2}{3\left(a^4+a^2+1\right)}=\dfrac{a^4+a^2+1-\left(a^4-2a^2+1\right)}{3\left(a^4+a^2+1\right)}=\dfrac{1}{3}-\dfrac{\left(a^2-1\right)^2}{3\left(a^4+a^2+1\right)}\le\dfrac{1}{3}\)
\(P_{max}=\dfrac{1}{3}\) khi \(a^2=1\Rightarrow a=\pm1\)
Ta có \(3P=\dfrac{3a^2}{a^4+a^2+1}=\dfrac{-a^4+2a^2-1+a^4+a^2+1}{a^4+a^2+1}=1-\dfrac{\left(a^2-1\right)^2}{a^4+a^2+1}\le1\)\(\Rightarrow P\le\dfrac{1}{3}\)
Dấu "=" xảy ra <=> a2 - 1 = 0 <=> a = \(\pm1\)
Vậy Max P = 1/3 khi a = \(\pm1\)
+) Dễ thấy \(P=\dfrac{a^2}{a^4+a^2+1}\ge0\) ("=" khi a = 0)
Vậy \(0\le P\le\dfrac{1}{3}\)
