Bài 1:

a: \(A=x^2+2x+4\)

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

\(=\left(x+1\right)^2+3>=3\forall x\)

Dấu '=' xảy ra khi x+1=0

=>x=-1

Vậy: \(A_{min}=3\) khi x=-1

b: \(B=x^2-20x+101\)

\(=x^2-20x+100+1\)

\(=\left(x-10\right)^2+1>=1\forall x\)

Dấu '=' xảy ra khi x-10=0

=>x=10

Vậy: \(B_{min}=1\) khi x=10

c: \(C=x^2-2x+y^2+4y+8\)

\(=x^2-2x+1+y^2+4y+4+3\)

\(=\left(x-1\right)^2+\left(y+2\right)^2+3>=3\forall x\)

Dấu '=' xảy ra khi x-1=0 và y+2=0

=>x=1 và y=-2

Vậy: \(C_{min}=3\) khi (x,y)=(1;-2)

Bài 2:

a: \(A=5-8x-x^2\)

\(=-\left(x^2+8x\right)+5\)

\(=-\left(x^2+8x+16-16\right)+5\)

\(=-\left(x+4\right)^2+16+5=-\left(x+4\right)^2+21< =21\forall x\)

Dấu '=' xảy ra khi x+4=0

=>x=-4

b: \(B=x-x^2\)

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

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

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

Dấu '=' xảy ra khi \(x-\dfrac{1}{2}=0\)

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

c: \(C=4x-x^2+3\)

\(=-x^2+4x-4+7\)

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

\(=-\left(x-2\right)^2+7< =7\forall x\)

Dấu '=' xảy ra khi x-2=0

=>x=2

d: \(D=-x^2+6x-11\)

\(=-\left(x^2-6x+11\right)\)

\(=-\left(x^2-6x+9+2\right)\)

\(=-\left(x-3\right)^2-2< =-2\forall x\)

Dấu '=' xảy ra khi x-3=0

=>x=3

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

Bài 6:

a) Ta có: \(A=-x^2+6x-11\)

\(=-\left(x^2-6x+11\right)\)

\(=-\left(x-3\right)^2-2\le-2\forall x\)

Dấu '=' xảy ra khi x=3

b) Ta có: \(B=-x^2-8x+5\)

\(=-\left(x^2+8x-5\right)\)

\(=-\left(x^2+8x+16-21\right)\)

\(=-\left(x+4\right)^2+21\le21\forall x\)

Dấu '=' xảy ra khi x=-4

c) Ta có: \(C=-x^2+4x+1\)

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

\(=-\left(x^2-4x+4-5\right)\)

\(=-\left(x-2\right)^2+5\le5\forall x\)

Dấu '=' xảy ra khi x=2

Bài 7:

a) Ta có: \(x^2-6x+11\)

\(=x^2-6x+9+2\)

\(=\left(x-3\right)^2+2\ge2\forall x\)

Dấu '=' xảy ra khi x=3

\(A=\left(x^2-4x+4\right)+4=\left(x-2\right)^2+4\ge4\)

\(minA=4\Leftrightarrow x=2\)

\(B=\left(4x^2-12x+9\right)+2=\left(2x-3\right)^2+2\ge2\)

\(minB=2\Leftrightarrow x=\dfrac{3}{2}\)

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

\(minC=-8\Leftrightarrow x=-1\)

\(D=-\left(x^2-2x+1\right)-4=-\left(x-1\right)^2-4\le-4\)

\(maxD=-4\Leftrightarrow x=1\)

\(E=-\left(4x^2-6x+\dfrac{9}{4}\right)-\dfrac{11}{4}=-\left(2x-\dfrac{3}{2}\right)^2-\dfrac{11}{4}\le-\dfrac{11}{4}\)

\(maxA=-\dfrac{11}{4}\Leftrightarrow x=\dfrac{3}{4}\)

\(F=-2\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)-\dfrac{55}{8}=-2\left(x-\dfrac{1}{4}\right)^2-\dfrac{55}{8}\le-\dfrac{55}{8}\)

\(maxF=-\dfrac{55}{8}\Leftrightarrow x=\dfrac{1}{4}\)

\(G=\left(x^2-4xy+4y^2\right)+\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-2y\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(maxG=\dfrac{3}{4}\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-\dfrac{1}{2}\end{matrix}\right.\)

\(H=-\left(x^2-2x+1\right)-\left(y^2+4y+4\right)+16=-\left(x-1\right)^2-\left(y+2\right)^2+16\le16\)

\(maxH=16\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

\(A=\left(x^2-2x+1\right)+4=\left(x-1\right)^2+4\ge4\\ A_{min}=4\Leftrightarrow x=1\\ B=2\left(x^2-3x\right)=2\left(x^2-2\cdot\dfrac{3}{2}x+\dfrac{9}{4}\right)-\dfrac{9}{2}\\ B=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\\ B_{min}=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{3}{2}\\ C=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\\ C_{max}=7\Leftrightarrow x=2\)

a,\(A=x^2-2x+5=\left(x^2-2x+1\right)+4=\left(x-1\right)^2+4\ge4\)

Dấu "=" \(\Leftrightarrow x=-1\)

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

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

c,\(=C=-\left(x^2-4x-3\right)=-\left[\left(x^2-4x+4\right)-7\right]=-\left(x-2\right)^2+7\le7\)

Dấu "=" \(\Leftrightarrow x=2\)

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

A = x² + 4x + 7

   = ( x² + 4x + 4 ) + 3

   = ( x + 2 )² + 3

( x + 2 )² ≥ 0 ∀ x => ( x + 2 )² + 3 ≥ 3

Đẳng thức xảy ra <=> x + 2 = 0 => x = -2

=> MinA = 3 <=> x = -2

B = 2x² - 6x 

   = 2( x² - 3x + 9/4 ) - 9/2

   = 2( x - 3/2 )² - 9/2

2( x - 3/2 )² ≥ 0 ∀ x => 2( x - 3/2 )² -9/2 ≥ -9/2 

Đẳng thức xảy ra <=> x - 3/2 = 0 => x = 3/2

=> MinB = -9/2 <=> x = 3/2

C = -2x² + 8x - 15

    = -2( x² - 4x + 4 ) - 7

    = -2( x - 2 )² - 7

-2( x - 2 )² ≤ 0 ∀ x => -2( x - 2 )² - 7 ≤ -7

Đẳng thức xảy ra <=> x - 2 = 0 => x = 2

=> MaxC = -7 <=> x = 2

Mình làm ở bài trước rồi nhé -..-

câu a khác mak

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

   \(=x^2-2x2+2^2+3\)

   \(=\left(x-2\right)^2+3\ge3\forall x\)

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

Vậy \(Min_A=3\Leftrightarrow x=2\)

\(B=2x^2-6x\)

    \(=2\left(x^2-3x\right)\)

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

   \(=2\left[\left(x-\frac{3}{2}\right)^2-\frac{9}{2}\right]\ge-\frac{9}{2}\forall x\)

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

Vậy \(Min_B=\frac{-9}{2}\Leftrightarrow x=\frac{3}{2}\)

\(C=-2x^2+8x-15\)

   \(=-2\left(x^2-4x+4\right)-7\)

    \(=-2\left(x-2\right)^2-7\le-7\forall x\)

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

Vậy \(Max_C=-7\Leftrightarrow x=2\)

\(a.A=\left(x-2\right)^2+\left(y+1\right)^2+1\ge1\forall x;y\) . " = " \(\Leftrightarrow x=2;y=-1\) 

b.\(B=7-\left(x+3\right)^2\le7\forall x\)  " = " \(\Leftrightarrow x=-3\)

c.\(C=\left|2x-3\right|-13\ge-13\forall x\)  " = " \(\Leftrightarrow x=\dfrac{3}{2}\)

d.\(D=11-\left|2x-13\right|\le11\forall x\)  " = " \(\Leftrightarrow x=\dfrac{13}{2}\)