\(A=-\left|3x-3\right|-\left(4x-4\right)^2-11\le-11\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}3x-3=0\\4x-4=0\end{matrix}\right.\Leftrightarrow x=1\)

5-/3x-4/

ta có: /3x-4/\(\ge0,\forall x\)

\(\Rightarrow\)5-/3x-4/\(\le5\)

Dấu "=" xảy ra khi 3x-4=0 =>3x=4 =>\(x=\frac{3}{4}\)

Vậy GTNL của 5-/3x-4/ là 5 với x=\(\frac{3}{4}\)

\(\left(4x-6\right)^{2008}+8\)

ta có: \(\left(4x-6\right)^{2008}\ge0,\forall x\)

\(\Rightarrow\left(4x-6\right)^{2008}+8\ge8\)

dấu "=" xảy ra khi (4x-6)²⁰⁰⁸=0

                           => 4x-6=0 =>4x=6 =>x=\(\frac{3}{2}\)

vậy GTNN của (4x-6)²⁰⁰⁸ là 8 với x=\(\frac{3}{2}\)

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

Do \(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\Rightarrow\dfrac{1}{x^2+x+1}\le\dfrac{4}{3}\)

\(\Rightarrow A\le3+\dfrac{4}{3}=\dfrac{13}{3}\)

\(maxA=\dfrac{13}{3}\Leftrightarrow x=-\dfrac{1}{2}\)

Ta có:\(\dfrac{3x^2+3x+4}{x^2+x+1}=\dfrac{3\left(x^2+x+1\right)+1}{x^2+x+1}=3+\dfrac{1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\)

Vì \(\left(x+\dfrac{1}{2}\right)^2\ge0\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge0\Leftrightarrow\dfrac{1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{4}{3}\)

\(\Rightarrow A\le3+\dfrac{4}{3}=\dfrac{13}{3}\)

Dấu "=" xảy ra \(\Leftrightarrow x=-\dfrac{1}{2}\)

là100

Bạn trình bày ra hộ mình nha

1, \(3x^2-5x+4\)

\(=3\left(x^2-\frac{5}{3}x\right)+1=3\left(x^2-2.\frac{5}{6}x+\frac{25}{36}\right)+\frac{23}{12}=3\left(x-\frac{5}{6}\right)^2+\frac{23}{12}\)

Ta có: \(3\left(x-\frac{5}{6}\right)^2\ge0\forall x\Leftrightarrow3\left(x-\frac{5}{6}\right)^2+\frac{23}{12}\ge\frac{23}{12}\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-\frac{5}{6}\right)^2=0\Leftrightarrow x-\frac{5}{6}=0\Leftrightarrow x=\frac{5}{6}\)

Vậy minA = \(\frac{23}{12}\Leftrightarrow x=\frac{5}{6}\)

2, Bạn thử kiểm tra lại đề bài xem