\(A=\dfrac{2x^2}{x^4+x^2+1}=\dfrac{6x^2}{3\left(x^4+x^2+1\right)}=\dfrac{2\left(x^4+x^2+1\right)-2x^4+4x^2-2}{3\left(x^4+x^2+1\right)}\)

\(A=\dfrac{2}{3}-\dfrac{2\left(x^2-1\right)^2}{3\left(x^4+x^2+1\right)}\le\dfrac{2}{3}\)

\(A_{max}=\dfrac{2}{3}\) khi \(x^2=1\)

Bạn ơi đề là M = \(\dfrac{x^2+x+1}{x^2+4}\) hay M = \(\dfrac{x^2+x+1}{x^2}+4\) vậy bn?

\(A=x^2-6x+10\)

\(\Leftrightarrow A=x^2-2\cdot x\cdot3+3^2-9+10\)

\(\Leftrightarrow A=\left(x-3\right)^2+1\ge1\)     \(\forall x\in z\)

\(\Leftrightarrow A_{min}=1khix=3\)

\(B=3x^2-12x+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x\right)^2-2\cdot\sqrt{3}x\cdot2\sqrt{3}+\left(2\sqrt{3}\right)^2-12+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x-2\sqrt{3}\right)^2-11\ge-11\)    \(\forall x\in z\)

\(\Leftrightarrow B_{min}=-11khix=2\)

Đặt P(x)=0

=>x²(-3x+2)=0

=>x=0 hoặc x=2/3

\(P\left(x\right)=x^2\cdot\left(2-3x\right)\)

Đặt \(P\left(x\right)=0\)

\(x^2\left(2-3x\right)=0\)

\(\Rightarrow\)\(-3x^3+2x^2=0\)

\(\Rightarrow\)\(x^2\left(-3x+2\right)=0\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x^2=0\\-3x+2=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy nghiệm của đa thức \(P\left(x\right)\) là \(x=\left\{0;\dfrac{2}{3}\right\}\)

\(A=\frac{5x^2+4x-1}{x^2}=\frac{9x^2-\left(4x^2-4x+1\right)}{x^2}=9-\frac{\left(2x-1\right)^2}{x^2}\le9\)

Dấu \(=\)khi \(2x-1=0\Leftrightarrow x=\frac{1}{2}\).

\(B=\frac{x^2}{x^2+x+1}=\frac{3x^2}{3x^2+3x+3}=\frac{4x^2+4x+4-\left(x^2+4x+4\right)}{3x^2+3x+3}=\frac{4}{3}-\frac{\left(x+2\right)^2}{3\left(x^2+x+1\right)}\le\frac{4}{3}\)

Dấu \(=\)khi \(x+2=0\Leftrightarrow x=-2\).

Bạn coi lại đề bài, mẫu số đoạn \(x^2+1a\) là sao nhỉ?