Có \(2x-2\sqrt{3x+1}-1\)

    \(=\left(2x+\frac{2}{3}\right)-2\sqrt{\left(2x+\frac{2}{3}\right).\frac{3}{2}}+\frac{3}{2}-\frac{19}{6}\)

     \(=\left(\sqrt{2x+\frac{2}{3}}-\sqrt{\frac{3}{2}}\right)^2-\frac{19}{6}\ge-\frac{19}{6}\forall x\ge-\frac{1}{3}\)

Dấu " =" xảy ra\(\Leftrightarrow\hept{\begin{cases}\sqrt{2x+\frac{2}{3}}=\sqrt{\frac{3}{2}}\\x\ge-\frac{1}{3}\end{cases}}\Leftrightarrow x=\frac{5}{12}\)

    Vậy....

\(A=\dfrac{2x+1}{x^2+2}\)

\(\Leftrightarrow Ax^{2\:}+2A=2x+1\)

+) \(A=0\Rightarrow x=-\dfrac{1}{2}\)

+) \(A\ne0\)

\(Ax^2+2A=2x+1\)

\(\Leftrightarrow Ax^{2\:}-2x=1-2A\)

\(\Leftrightarrow x^2-2.\dfrac{x}{A}=\dfrac{1-2A}{A}\)

\(\Leftrightarrow x^2-2.x.\dfrac{1}{A}+\dfrac{1}{A^2}=\dfrac{1-2A}{A}+\dfrac{1}{A^2}\)

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

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

Vì \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{A}\right)^2\ge0\left(\forall x,A\ne0\right)\\A^2\ge0\end{matrix}\right.\)

⇒ \(\left(1-A\right)\left(2A+1\right)\ge0\)

⇒ \(-\dfrac{1}{2}\le A\le1\)

Còn lại tụ làm nha

\(A=\dfrac{2x+1}{x^2+2}=\dfrac{x^2+2-x^2-2+2x+1}{x^2+2}\\ =1-\dfrac{-\left(x-1\right)^2}{x^2+2}\\ Do\left(x-1\right)^2\ge0\Rightarrow\dfrac{-\left(x-1\right)^2}{x^2+2}\ge0\\ \Rightarrow\dfrac{-\left(x-1\right)^2}{x^2+2}=0\Leftrightarrow\dfrac{-\left(x-1\right)^2}{x^2+2}+1\le1\) 

\(Dấu"="\Leftrightarrow A=1\\ \Leftrightarrow x-1=0\Rightarrow x=1\\ Vậy.P_{max}=1.khi.x=1\\ A=\dfrac{2x+1}{x^2+2}\rightarrow2A+1=\dfrac{2.\left(2x+1\right)}{x^2+2}+1\\ =\dfrac{4x+2+x^2+2}{x^2+2}=\dfrac{x^2+4x+2}{x^2+2}=\dfrac{\left(x+2\right)^2}{x^2+2}\\ Do\left(x+2\right)^2\ge0\Leftrightarrow\dfrac{\left(x+2\right)^2}{x^2+2}\ge0\) 

\(Dấu"="\Leftrightarrow A=\dfrac{1}{2}khi.x=-2\\ \Rightarrow2A+1\ge0\Rightarrow2A\ge-1\Rightarrow A>-\dfrac{1}{2}\\ Vậy.MinA=-\dfrac{1}{2}.khi.x=-2\)

A=2x² + 3x + 1 

A=2(x² + 3/2x + 9/16) -1/8

A=2(x+3/4)² - 1/8

Vậy GTNN của A là -1/8

\(A=3\left(x^2-\frac{2}{3}x-\frac{1}{3}\right)\)

\(A=3\left(x^2-2\cdot\frac{1}{3}x+\frac{1}{9}-\frac{4}{9}\right)\)

\(A=\left(x-\frac{1}{3}\right)^2-\frac{4}{3}\)\(\supseteq-\frac{4}{3}\)

Dấu = xr khi x=1/3

Vậy Min A=-4/3 tại x=1/3

\(A=3x^2-2x-1\)

\(=3\left(x^2-\frac{2}{3}x-\frac{1}{3}\right)\)

\(=3\left(x^2-2.x.\frac{1}{3}+\frac{1}{9}-\frac{1}{9}-\frac{1}{3}\right)\)

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

Vì \(3\left(x-\frac{1}{3}\right)^2\ge0;\forall x\)

\(\Rightarrow3\left(x-\frac{1}{3}\right)^2-\frac{4}{3}\ge0-\frac{4}{3};\forall x\)

Hay \(A\ge\frac{-4}{3};\forall x\)

Dấu"=" xảy ra \(\Leftrightarrow\left(x-\frac{1}{3}\right)^2=0\)

                      \(\Leftrightarrow x=\frac{1}{3}\)

Vậy MIN \(A=\frac{-4}{3}\)\(\Leftrightarrow x=\frac{1}{3}\)

- 1 hay + 1 v?

+ 1 hợp lí hơn

\(A=3x^2-2x-1\)

\(A=3\left(x^2-\frac{2}{3}x-\frac{1}{3}\right)\)

\(A=3\left(x^2-2.\frac{1}{3}.x+\frac{1}{9}-\frac{4}{9}\right)\)

\(A=3\left(x-\frac{1}{3}\right)^2-\frac{4}{3}\ge-\frac{4}{3}\)với \(\forall x\)

Dấu "=" xảy ra <=> x=1/3

Vậy GTNN của A là -4/3 <=> x=1/3

đặt x^2-7x=y=> \(y\ge-\frac{49}{4}\) (*)

\(A=y\left(y+12\right)=y^2+12y=\left(y+6\right)^2-36\ge-36\)

đẳng thức khi y=-6 thủa mãn đk (*)

Vậy: GTNN của A=-36 khí y=-6 =>\(\left[\begin{matrix}x=1\\x=6\end{matrix}\right.\)