Tìm min:

$F=3x^2+x-2=3(x^2+\frac{x}{3})-2$

$=3[x^2+\frac{x}{3}+(\frac{1}{6})^2]-\frac{25}{12}$

$=3(x+\frac{1}{6})^2-\frac{25}{12}\geq \frac{-25}{12}$

Vậy $F_{\min}=\frac{-25}{12}$. Giá trị này đạt tại $x+\frac{1}{6}=0$
$\Leftrightarrow x=\frac{-1}{6}$

Tìm min

$G=4x^2+2x-1=(2x)^2+2.2x.\frac{1}{2}+(\frac{1}{2})^2-\frac{5}{4}$

$=(2x+\frac{1}{2})^2-\frac{5}{4}\geq 0-\frac{5}{4}=\frac{-5}{4}$ (do $(2x+\frac{1}{2})^2\geq 0$ với mọi $x$)

Vậy $G_{\min}=\frac{-5}{4}$. Giá trị này đạt tại $2x+\frac{1}{2}=0$

$\Leftrightarrow x=\frac{-1}{4}$

Tìm min

$H=5x^2-x+1=5(x^2-\frac{x}{5})+1$

$=5[x^2-\frac{x}{5}+(\frac{1}{10})^2]+\frac{19}{20}$

$=5(x-\frac{1}{10})^2+\frac{19}{20}\geq \frac{19}{20}$
Vậy $H_{\min}=\frac{19}{20}$. Giá trị này đạt tại $x-\frac{1}{10}=0$

$\Leftrightarrow x=\frac{1}{10}$

BÀI 1:

a)  \(A=x^2+2x+3=\left(x+1\right)^2+2\ge2\)

Vậy MIN   \(A=2\)khi \(x=-1\)

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

Vậy MIN   \(B=\frac{3}{4}\)khi    \(x=-\frac{1}{2}\)

c)  \(C=2x^2+3x-1=2\left(x+\frac{3}{4}\right)^2-\frac{17}{8}\ge-\frac{17}{8}\) 

Vậy MIN   \(C=-\frac{17}{8}\)khi     \(x=-\frac{3}{4}\)

d)  \(D=4x^2-x=\left(2x-\frac{1}{4}\right)^2-\frac{1}{16}\ge-\frac{1}{16}\)

Vậy  MIN  \(D=-\frac{1}{16}\)khi    \(x=\frac{1}{8}\)

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

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

\(=3\left(x^2+x+\dfrac{1}{4}-\dfrac{7}{12}\right)\)

\(=3\left(x+\dfrac{1}{2}\right)^2-\dfrac{7}{4}>=-\dfrac{7}{4}\forall x\)

Dấu '=' xảy ra khi x+1/2=0

=>\(x=-\dfrac{1}{2}\)

\(C=-2x^2+7x+3\)

\(=-2\left(x^2-\dfrac{7}{2}x-\dfrac{3}{2}\right)\)

\(=-2\left(x^2-2\cdot x\cdot\dfrac{7}{4}+\dfrac{49}{16}-\dfrac{73}{16}\right)\)

\(=-2\left(x-\dfrac{7}{4}\right)^2+\dfrac{73}{8}< =\dfrac{73}{8}\forall x\)

Dấu '=' xảy ra khi x-7/4=0

=>x=7/4

a) \(A=\sqrt{x-2}+\sqrt{6-x}\)

\(\Rightarrow A^2=x-2+6-x+2\sqrt{\left(x-2\right)\left(6-x\right)}\)

Ta có \(\sqrt{\left(x-2\right)\left(6-x\right)}\ge0,\forall x\)

Do đó \(A^2=4+2\sqrt{\left(x-2\right)\left(6-x\right)}\ge4\)

Mà A không âm \(\Leftrightarrow A\ge2\)

Dấu "=" \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

Áp dụng BĐT Bunhiacopxky:

\(A^2=\left(\sqrt{x-2}+\sqrt{6-x}\right)^2\le\left(x-2+6-x\right)\left(1+1\right)=4\cdot2=8\)

\(\Leftrightarrow A\le\sqrt{8}\)

Dấu "=" \(\Leftrightarrow x-2=6-x\Leftrightarrow x=4\)

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