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

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

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

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

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

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

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

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

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

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

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

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?

Xét \(P^2=2+\sqrt{\left(x-2\right)\left(4-x\right)}\)

Áp dụng bất đẳng thức \(2\sqrt{ab}\le a+b\)ta có

                \(P^2\le2+\left(x-2\right)+\left(4-x\right)=4\)

Từ đó  max \(P=2\Leftrightarrow x-2=4-x\Leftrightarrow x=3\)

cảm ơn bạn nhiều nha!!!!!!

Bài làm:

Ta có: \(\left|x-4\right|.\left(2-\left|x-4\right|\right)\)

\(=-\left|x-4\right|^2+2.\left|x-4\right|\)

\(=-\left(\left|x-4\right|^2-2.\left|x-4\right|+1\right)+1\)

\(=-\left(\left|x-4\right|-1\right)^2+1\le1\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(-\left(\left|x-4\right|-1\right)^2=0\Leftrightarrow\left|x-4\right|=1\Leftrightarrow\orbr{\begin{cases}x=3\\x=5\end{cases}}\)

Vậy Max = 1 khi x = 3 hoặc x = 5

có cái bài dễ vậy ko làm dc

\(\left|x-4\right|\left(2-\left|x-4\right|\right)\)

+) Nếu \(x\ge4\)

\(\left|x-4\right|\left(2-\left|x-4\right|\right)=\left(x-4\right)\left(6-x\right)=6x-x^2-24+4x\)

\(=x^2-10x-24=-\left(x-5\right)^2+1\le1\)

Dấu "=" xảy ra \(\Leftrightarrow-\left(x-5\right)^2=0\Leftrightarrow x-5=0\Leftrightarrow x=5\) ( tmđk )

+) Nếu \(x< 4\)

\(\left|x-4\right|\left(2-\left|x-4\right|\right)=\left(-x+4\right)\left(2+x-4\right)=\left(4-x\right)\left(x-2\right)\)

\(=4x-8-x^2+2x=-x^2+6x-8=-\left(x-3\right)^2+1\le1\)

Dấu "=" xảy ra \(\Leftrightarrow-\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\) ( tmđk )

Vậy GTLN của bt trên = 1 \(\Leftrightarrow\orbr{\begin{cases}x=5\\x=3\end{cases}}\)