\(A=x^2-4x+10=x^2-4x+4+6=\left(x-2\right)^2+6\ge6\)

Vậy GTNN A là 6 khi x - 2 = 0 <=> x = 2 

\(B=\left(1-x\right)\left(3x-4\right)=3x-4-3x^2+4x=-3x^2+7x-4\)

\(=-3\left(x^2-\frac{7}{3}x+\frac{4}{3}\right)=-3\left(x^2-2.\frac{7}{6}x+\frac{49}{36}-\frac{1}{36}\right)=-3\left(x-\frac{7}{6}\right)^2+\frac{1}{12}\ge\frac{1}{12}\)

\(=3\left(x-\frac{7}{6}\right)^2-\frac{1}{12}\le-\frac{1}{12}\)Vậy GTLN B là -1/12 khi x = 7/6 

\(C=3x^2-9x+5=3\left(x^2-3x+\frac{5}{3}\right)=3\left(x^2-2.\frac{3}{2}x+\frac{9}{4}-\frac{7}{12}\right)\)

\(=3\left(x-\frac{3}{2}\right)^2-\frac{7}{4}\ge-\frac{7}{4}\)Vậy GTNN C là -7/4 khi x = 3/2 

\(D=-2x^2+5x+2=-2\left(x^2-\frac{5}{2}x-1\right)=-2\left(x^2-2.\frac{5}{4}x+\frac{25}{16}-\frac{41}{16}\right)\)

\(=-2\left(x-\frac{5}{4}\right)^2+\frac{21}{8}\le\frac{21}{8}\)Vậy GTLN D là 21/8 khi x = 5/4 

\(A=2\left(x^2-4x+4\right)-7=2\left(x-2\right)^2-7\ge-7\)

Dấu \("="\Leftrightarrow x=2\)

\(B=\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{1}{4}=\left(x+\dfrac{3}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

Dấu \("="\Leftrightarrow x=-\dfrac{3}{2}\)

\(C=4\left(x^2-2x+1\right)-4=4\left(x-1\right)^2-4\ge-4\)

Dấu \("="\Leftrightarrow x=1\)

\(D=\dfrac{1}{-\left(x^2+2x+1\right)+6}=\dfrac{1}{-\left(x+1\right)^2+6}\ge\dfrac{1}{6}\)

Dấu \("="\Leftrightarrow x=-1\)

\(A=2\left(x^2-4x+4\right)-7=2\left(x-2\right)^2-7\ge-7\)

\(A_{min}=-7\) khi \(x=2\)

\(B=\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{1}{4}=\left(x+\dfrac{3}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

\(B_{min}=-\dfrac{1}{4}\) khi \(x=-\dfrac{3}{2}\)

\(C=4\left(x^2-2x+1\right)-4=4\left(x-1\right)^2-4\ge-4\)

\(C_{min}=-4\) khi \(x=1\)

Biểu thức D không tồn tại cả max lẫn min

1.

$A=2x^2-8x+1=2(x^2-4x+4)-7=2(x-2)^2-7$

Vì $(x-2)^2\geq 0$ với mọi $x\in\mathbb{R}$

$\Rightarrow A\geq 2.0-7=-7$

Vậy $A_{\min}=-7$ khi $x-2=0\Leftrightarrow x=2$

2.

$B=x^2+3x+2=(x^2+3x+1,5^2)-0,25=(x+1,5)^2-0,25\geq 0-0,25=-0,25$

Vậy $B_{\min}=-0,25$ khi $x=-1,5$

3.

$C=4x^2-8x=(4x^2-8x+4)-4=(2x-2)^2-4\geq 0-4=-4$

Vậy $C_{\min}=-4$ khi $2x-2=0\Leftrightarrow x=1$

4. Để $D_{\min}$ thì $5-x^2-2x$ là số thực âm lớn nhất

Mà không tồn tại số thực âm lớn nhất nên không tồn tại $x$ để $D_{\min}$

A\(=2x^2-8x+1\)

=2x(x-4)+1≥1

Min A=1 ⇔x=4

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

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

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

Min B=-1/4⇔x=-3/2

C=\(4x^2-8x\)

=\(\left(\left(2x\right)^2-2x.4+16\right)-16\)

=(2x-4)^2 -16≥-16

Min C=-16 ⇔x=2

D=\(\dfrac{1}{-\left(x^2-2x+1\right)+6}\)

=\(\dfrac{1}{-\left(x-1\right)^2+6}\)≥\(\dfrac{1}{6}\)

Min D=1/6 ⇔x=1

a. Ta có : \(A=\frac{8x^2-9}{x^2+3}=\frac{8x^2+24-33}{x^2+3}=8-\frac{33}{x^2+3}\)

Để Amin thì \(\frac{33}{x^2+3}_{max}\) mà \(\frac{33}{x^2+3}\le11\)

Dấu "=" xảy ra \(\Leftrightarrow x^2+3=3\Leftrightarrow x=0\)

Vậy Amin = 8 - 11 = - 3 <=> x = 0

b. Ta có : \(B=\frac{3x^2-6x+40}{x^2-2x+5}=\frac{3\left(x^2-2x+5\right)+25}{x^2-2x+5}=3+\frac{25}{x^2-2x+5}\)

Để Bmax thì \(\frac{25}{x^2-2x+5}=\frac{25}{\left(x-1\right)^2+4}_{max}\)

mà \(\frac{25}{\left(x-1\right)^2+4}\le\frac{25}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-1\right)^2+4=4\Leftrightarrow x-1=0\Leftrightarrow x=1\)

Vậy Bmax \(=3+\frac{25}{4}=\frac{37}{4}\)  <=> x = 1

a) \(A=x^2-10x+5\)

\(A=x^2-10x+25-20\)

\(A=\left(x-5\right)^2-20\ge-20\)

Min A = -20 \(\Leftrightarrow x=5\)

b) \(B=3x^2-6x+11\)

\(B=3\left(x^2-2x+1\right)+8\)

\(B=3\left(x-1\right)^2+8\ge8\)

Min B = 8\(\Leftrightarrow x=1\)

a) \(A=x^2-10x+5=\left(x^2-10x+25\right)-20\)

\(=\left(x-5\right)^2-20\ge-20\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left(x-5\right)^2=0\Rightarrow x=5\)

Vậy \(Min_A=-20\Leftrightarrow x=5\)

b) \(B=3x^2-6x+11=3\left(x^2-2x+1\right)+8\)

\(=3\left(x-1\right)^2+8\ge8\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left(x-1\right)^2=0\Rightarrow x=1\)

Vậy \(Min_B=8\Leftrightarrow x=1\)

c) \(C=8x^2+10x-30=8\left(x^2-\frac{5}{4}x+\frac{25}{64}\right)-\frac{265}{8}\)

\(=8\left(x-\frac{5}{8}\right)^2-\frac{265}{8}\ge-\frac{265}{8}\)

Dấu "=" xảy ra khi: \(\left(x-\frac{5}{8}\right)^2=0\Rightarrow x=\frac{5}{8}\)

Vậy \(Min_C=-\frac{265}{8}\Leftrightarrow x=\frac{5}{8}\)

A = x² - 10x + 5

A = ( x² - 10x + 25 ) - 20

A = ( x - 5 )² - 20

( x - 5 )² ≥ 0 ∀ x => ( x - 5 )² - 20 ≥ -20

Đẳng thức xảy ra <=> x - 5 = 0 => x = 5

=> MinA = -20 <=> x = 5

b) B = 3x² - 6x + 11

B = 3( x² - 2x + 1 ) + 8

B = 3( x - 1 )² + 8

3( x - 1 )² ≥ 0 ∀ x => 3( x - 1 )² + 8 ≥ 8

Đẳng thức xảy ra <=> x - 1 = 0 => x = 1

=> MinB = 8 <=> x = 1

C = 8x² + 10x - 30

C = 8( x² + 5/4x + 25/64 ) - 265/8

C = 8( x + 5/8 )² - 265/8

8( x + 5/8 )² ≥ 0 ∀ x => 8( x + 5/8 )² - 265/8 ≥ -265/8

Đẳng thức xảy ra <=> x + 5/8 = 0 => x = -5/8

=> MinC = -265/8 <=> x = -5/8

c: \(-x^2+2x-2=-\left(x-1\right)^2-1\le-1\forall x\)

\(\Leftrightarrow V\ge-1\forall x\)

Dấu '=' xảy ra khi x=1

a: Ta có: \(x^2+x+1\)

\(=x^2+2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

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

Dấu '=' xảy ra khi \(x=-\dfrac{1}{2}\)

b: Ta có: \(-x^2+x+2\)

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

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

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

a)  \(N=13-6x-3x^2\)

\(-N=3x^2+6x-13\)

\(-N=3\left(x^2+2x+1\right)-16\)

\(-N=3\left(x+1\right)^2-16\)

Mà  \(\left(x+1\right)^2\ge0\forall x\)\(\Rightarrow3\left(x+1\right)^2\ge0\forall x\)

\(\Rightarrow-N\ge-16\)

\(\Leftrightarrow N\le16\)

Dấu "=" xảy ra khi :  \(x+1=0\Leftrightarrow x=-1\)

Vậy ...

b)  \(Q=2x^2-8x+11\)

\(Q=2\left(x^2-4x+4\right)+3\)

\(Q=2\left(x-2\right)^2+3\)

Mà  \(\left(x-2\right)^2\ge0\forall x\) \(\Rightarrow2\left(x-2\right)^2\ge0\forall x\)

\(\Rightarrow Q\ge3\)

Dấu "=" xảy ra khi :  \(x-2=0\Leftrightarrow x=2\)

Vậy ...

ran love shinichi................

CẠN CMN LỜI