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 $x-\frac{1}{10}=0$

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

\(\frac{5}{4x^2-4x+21}=\frac{5}{4x^2-2x-2x+1+20}=\frac{5}{\left(2x-1\right)^2+20}\)

\(\left(2x-1\right)^2\ge0\Rightarrow\left(2x-1\right)^2+20\ge20\)

dấu = xảy ra khi (2x-1)²=0

=> \(x=\frac{1}{2}\)

Vậy max \(\frac{5}{4x^2-4x+21}=\frac{5}{20}=\frac{1}{4}\)

a) \(A=5x^2-6x-1\)

   \(\Rightarrow A=5\left(x^2-\frac{6}{5}x-\frac{1}{5}\right)\)

  \(\Rightarrow A=5\left(x^2-2\cdot x\cdot\frac{6}{10}+\frac{36}{100}-\frac{14}{25}\right)\)

  \(\Rightarrow A=5\left[\left(x-\frac{6}{10}\right)^2-\frac{14}{25}\right]\)

  \(\Rightarrow A=5\left(x-\frac{6}{10}\right)^2-\frac{14}{5}\)

  Vì \(\left(x-\frac{6}{10}\right)^2\ge0\forall x\)\(\Rightarrow A=5\left(x-\frac{6}{10}\right)^2-\frac{14}{5}\ge-\frac{14}{5}\forall x\)

\(A=-\frac{14}{5}\Leftrightarrow\left(x-\frac{6}{10}\right)^2=0\Leftrightarrow x=\frac{6}{10}\)

Vậy \(MinA=-\frac{14}{5}\Leftrightarrow x=\frac{6}{10}\)

\(x^2+y^2+2xy+4x+4y\)

\(=\left(x+y\right)^2+4\left(x+y\right)\)

\(=\left(x+y\right)\left(x+y+4\right)\)

Trần Minh Hoàng

Đề bảo tìm Min hoặc Max mà bn

Bài 1:

a. A = x^2 - 5x - 1

\(=x^2-5x+\frac{25}{4}-\frac{29}{4}\)

\(=x^2-5x+\left(\frac{5}{2}\right)^2-\frac{29}{4}\)

\(=\left(x-\frac{5}{2}\right)^2-\frac{29}{4}\ge0-\frac{29}{4}=-\frac{29}{4}\)

Dấu = khi x=5/2

Vậy MinC=-29/4 khi x=5/2

2. Tìm x:
a. ( 2x - 3 )^2 - ( 4x + 1 )( 4x - 1 ) = ( 2x - 1 ).( 3 - 7x )

=>4x²-12x+9+1-16x²=-14x²+13x-3

=>-12x²-12x+10=-14x²+13x-3

=>2x²-25x+13=0

\(\Rightarrow2\left(x-\frac{25}{4}\right)^2-\frac{521}{8}=0\)

\(\Rightarrow\left(x-\frac{25}{4}\right)^2=\frac{521}{16}\)

\(\Rightarrow x-\frac{25}{4}=\pm\sqrt{\frac{521}{16}}\)

\(\Rightarrow x=\frac{25}{4}\pm\frac{\sqrt{521}}{4}\)

c. 4.( x - 3 ) - ( x + 2 ) = 0

=>4x-12-x-2=0

=>3x-14=0

=>3x=14

=>x=14/3

a)\(-x^2-x+2\)

\(=-\left(x^2+x-2\right)\)

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

\(=-\left(x+\frac{1}{2}\right)^2+\frac{7}{4}\le\frac{7}{4}.Với\forall x\in Z\)

Dấu "="  xảy ra khi 

\(x+\frac{1}{2}=0\Leftrightarrow x=-\frac{1}{2}\)

Vậy Max = 7/4 <=> x = -1/2

a, \(-\left(x^2+x\right)+2\)

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

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

Dấu = xảy ra khi : \(x+\frac{1}{2}=0\)

=> \(x=-\frac{1}{2}\) 

Vậy Max=9/4 khi x=-1/2 

Các câu khác tương tự nhé.

A=1-[(√5x)^2+2.2(√5.x)/√5+4/5]+4/5

A=9/5-(√5x+2/√5)^2<=9/5

GtlnA=9/5 khi x=-2/5

a) A = -5x² - 4x + 1

A = -5( x² + 2.\(\dfrac{2}{5}\)x + \(\dfrac{4}{25}-\dfrac{4}{25}\)) + 1

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

A = -5\(\left(x+\dfrac{2}{5}\right)^2\) + \(\dfrac{9}{5}\)

Do : -5\(\left(x+\dfrac{2}{5}\right)^2\) ≥ 0 ∀x

⇒ -5\(\left(x+\dfrac{2}{5}\right)^2\) + \(\dfrac{9}{5}\) ≥ \(\dfrac{9}{5}\)

⇒ A_MAX = \(\dfrac{9}{5}\) ⇔ x = \(\dfrac{-2}{5}\)

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

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

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

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

Do : - \(\dfrac{\left(x-1\right)^2}{x^2+2}\) ≤ 0 ∀x

⇒ 1 - \(\dfrac{\left(x-1\right)^2}{x^2+2}\) ≤ 1

⇒ B_Max = 1 ⇔ x = 1