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 $2x+\frac{1}{2}=0$

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

Giusp e ạ !

a, \(3x^3-5x^2-x-2>0\)

\(< =>3x^3+x^2+x-6x^2-2x-2>0\)

\(< =>x\left(3x^2+x+1\right)-2\left(3x^2+x+1\right)>0\)

\(< =>\left(x-2\right)\left(3x^2+x+1\right)>0\)

có \(3x^2+x+1=3\left(x^2+\dfrac{1}{3}x+\dfrac{1}{3}\right)=3\left[x^2+2.\dfrac{1}{6}x+\dfrac{1}{36}+\dfrac{35}{36}\right]\)

\(=3\left[\left(x+\dfrac{1}{6}\right)^2+\dfrac{35}{36}\right]>0=>x-2>0< =>x>2\)

b, \(A=2x^2+y^2-2xy-2x+3\)

\(A=x^2-2xy+y^2+x^2-2x+1+2\)

\(A=\left(x-y\right)^2+\left(x-1\right)^2+2\ge2\)

dấu"=" xảy ra<=>\(x=y=1\)

( x - 3)( x - 5) + 4 = x^2 - 3x - 5x + 15 + 4 = x^2 - 8x + 19 = x^2 -8x + 16 + 3 = (x - 4)^2 + 3 

Vì( x + 4)^2 > = 0 với mọi x => ( x + 4)^2 + 3 lớn hơn bằng 3 

VẬy GTNN của bt là 3 khi x + 4 = 0 => x = - 4

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    \(x=\frac{1}{8}\)

giúp mình mọi người ơiii

Bài 1.

A = 2x² - x + 4 = 2( x² - 1/2x + 1/16 ) + 31/8 = 2( x - 1/4 )² + 31/8 ≥ 31/8 ∀ x

Dấu "=" xảy ra khi x = 1/4

=> MinA = 31/8 <=> x = 1/4

Bài 2.

A = -x² + 3x + 2 = -( x² - 3x + 9/4 ) + 17/4 = -( x - 3/2 )² + 17/4 ≤ 17/4 ∀ x

Dấu "=" xảy ra khi x = 3/2

=> MaxA = 17/4 <=> x = 3/2

B = 3x² + x - 5 = 3( x² + 1/3x + 1/36 ) - 61/12 = 3( x + 1/6 )² - 61/12 ≥ -61/12 ∀ x

Dấu "=" xảy ra khi x = -1/6

=> MinB = -61/12 <=> x = -1/6

C = x² + 3/2x - 5 = ( x² + 3/2x + 9/16 ) - 89/16 = ( x + 3/4 )² - 89/16 ≥ -89/16 ∀ x

Dấu "=" xảy ra khi x = -3/4

=> MinC = -89/16 <=> x= -3/4

Ta có : A = 9x² - 6x + 2 

= 9x² - 6x + 1 + 1 = (3x - 1)² + 1 \(\ge\)1 

=> Min A = 1

Dấu "=" xảy ra <=> 3x - 1 = 0 

<=> x = 1/3

Vậy Min A = 1 <=> x = 1/3

b) Ta có 2B = 4x² + 4x + 2 

= 4x² + 4x + 1 + 1 

= (2x + 1)² + 1 \(\ge\)1

=> B \(\ge\frac{1}{2}\)

Dấu "=" xảy ra <=> 2x + 1 = 0 

<=> x = -1/2

Vậy Min B = 1/2 <=> x = -1/2

c) C = (2x - 1)² + (x - 2)² 

= 5x² - 8x + 5

=> 5C = 25x² - 40x + 25 

 = 25x² - 40x + 16 + 9 

= (5x - 4)² + 9 \(\ge9\)

=> \(C\ge\frac{9}{5}\)

Dấu "=" xảy ra <=> 5x - 4 = 0 

<=> x = 0,8

Vậy Min C = 9/5 <=> x = 0,8

d) D = 3x² + 5x = \(3\left(x^2+\frac{5}{3}x\right)=3\left(x^2+2.\frac{5}{6}x+\frac{25}{36}-\frac{25}{36}\right)=3\left(x+\frac{5}{6}\right)^2-\frac{25}{12}\ge-\frac{25}{12}\)

=> \(D\ge-\frac{25}{12}\)

Dấu "=" xảy ra <=> x + 5/6 = 0 

<=> x = -5/6

Vậy Min D = -25/12 <=> x = -5/6e) E = (x -2)(x - 3)(x + 5)x

= (x² - 5x + 6)(x² + 5x)