\(MinA=0\Leftrightarrow7+4x-4x^2=0\Leftrightarrow x=\dfrac{1\pm2\sqrt{2}}{2}\)

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a: \(A=\left(x^2-2xy+y^2\right)+2x^2-7\)

\(=\left(x-y\right)^2+2x^2-7\ge-7\forall x,y\)

Dấu '=' xảy ra khi x=y=0

b: \(B=4x^2+4x+1-1=\left(2x+1\right)^2-1\ge-1\forall x\)

Dấu '=' xảy ra khi x=-1/2

Bài 2: 

a: Ta có: \(x^2+4x+7\)

\(=x^2+4x+4+3\)

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

Dấu '=' xảy ra khi x=-2

a.

\(A=\dfrac{2013}{x^2}-\dfrac{2}{x}+1=2013\left(\dfrac{1}{x}-\dfrac{1}{2013}\right)^2+\dfrac{2012}{2013}\ge\dfrac{2012}{2013}\)

Dấu "=" xảy ra khi \(x=2013\)

b.

\(B=\dfrac{4x^2+2-4x^2+4x-1}{4x^2+2}=1-\dfrac{\left(2x-1\right)^2}{4x^2+2}\le1\)

\(B_{max}=1\) khi \(x=\dfrac{1}{2}\)

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

\(B_{max}=-\dfrac{1}{2}\) khi \(x=-1\)

a)A=4(x+11/8)^2 -153/16

Min A=-153/16 khi x=-11/8

b)B=3(x-1/3)^2 -4/3

Min B=-4/3 khi x=1/3

Bài 1:

a) \(A=4x^2+11x-2=\left(4x^2+11x+\dfrac{121}{16}\right)-\dfrac{153}{16}=\left(2x+\dfrac{11}{4}\right)^2-\dfrac{153}{16}\ge-\dfrac{153}{16}\)

\(minA=-\dfrac{153}{16}\Leftrightarrow x=-\dfrac{11}{8}\)

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

\(minB=-\dfrac{4}{3}\Leftrightarrow x=\dfrac{1}{3}\)

Bài 2:

a) \(A=-x^2+3x-1=-\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{5}{4}=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{5}{4}\le\dfrac{5}{4}\)

\(maxA=\dfrac{5}{4}\Leftrightarrow x=\dfrac{3}{2}\)

b) \(B=-x^2-4x+7=-\left(x^2+4x+4\right)+11=-\left(x+2\right)^2+11\le11\)

\(maxB=11\Leftrightarrow x=-2\)

Bài 1: 

a: Ta có: \(A=4x^2+11x-2\)

\(=4\left(x^2+\dfrac{11}{4}x-\dfrac{1}{2}\right)\)

\(=4\left(x^2+2\cdot x\cdot\dfrac{11}{8}+\dfrac{121}{64}-\dfrac{153}{64}\right)\)

\(=4\left(x+\dfrac{11}{8}\right)^2-\dfrac{153}{16}\ge-\dfrac{153}{16}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{11}{8}\)

b: Ta có: \(B=3x^2-2x-1\)

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

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

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

Dấu '=' xảy ra khi \(x=\dfrac{1}{3}\)

`A=x^2-2x+5`

`=x^2-2x+1+4`

`=(x-1)^2+4>=4`

Dấu "=" `<=>x=1`

`B=4x^2+4x+3`

`=4x^2+4x+1+2`

`=(2x+1)^2+2>=2`

Dấu "=" xảy ra khi `x=-1/2`

`C=9x^2-6x+7`

`=9x^2-6x+1+6`

`=(3x-1)^2+6>=6`

Dấu '=' xảy ra khi `x=1/3`

`D=5x^2+3x+8`

`=5(x^2+3/5x)+8`

`=5(x^2+3/5x+9/100-9/100)+8`

`=5(x+3/10)^2+151/20>=151/20`

Dấu "=" xảy ra khi `x=-3/10`

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

Ta có: \(\left(x-1\right)^2\ge0\Rightarrow\left(x-1\right)^2+4\ge4\Rightarrow A_{min}=4\) khi \(x=1\)

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

Ta có: \(\left(2x+1\right)^2\ge0\Rightarrow\left(2x+1\right)^2+2\ge2\Rightarrow B_{min}=2\) khi \(x=-\dfrac{1}{2}\)

\(C=9x^2-6x+7=9x^2-6x+1+6=\left(3x-1\right)^2+6\)

Ta có: \(\left(3x-1\right)^2\ge0\Rightarrow\left(3x-1\right)^2+6\ge6\Rightarrow C_{min}=6\) khi \(x=\dfrac{1}{3}\)

\(D=5x^2+3x+8\Rightarrow5\left(x^2+2.x.\dfrac{3}{10}+\dfrac{9}{100}\right)+\dfrac{151}{20}=5\left(x+\dfrac{3}{10}\right)^2+\dfrac{151}{20}\)

Ta có: \(5\left(x+\dfrac{3}{10}\right)^2\ge0\Rightarrow5\left(x+\dfrac{3}{10}\right)^2+\dfrac{151}{20}\ge\dfrac{151}{20}\)

\(\Rightarrow D_{min}=\dfrac{151}{20}\) khi \(x=-\dfrac{3}{10}\)

- A = (x-1)² + 4 \(\ge4\)

Dấu "=" <=> x = 1

- B = (2x+1)² +2 \(\ge2\)

Dấu "=" xảy ra <=> x = \(\dfrac{-1}{2}\)

- C = (3x - 1)² + 6 \(\ge6\)

Dấu "=" <=> x = \(\dfrac{1}{3}\)

- D = \(5\left(x^2+\dfrac{3}{5}x+\dfrac{9}{100}\right)+\dfrac{151}{20}=5\left(x+\dfrac{3}{10}\right)^2+\dfrac{151}{20}\ge\dfrac{151}{20}\)

Dấu "=" <=> x = \(\dfrac{-3}{10}\)

a ) \(A=x^2-4x-7\)

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

     \(A=\left(x+2\right)^2-11\)

Ta có : \(\left(x+2\right)^2\ge0\)

  \(\Rightarrow\left(x+2\right)^2-11\ge-11\)

Vậy GTNN của \(A=-11\)

Khi : \(x+2=0\)

         \(x=-2\)

b ) \(B=-x^2+4x-7\)

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

     \(B=-\left(x-2\right)^2-3\)

Ta có : \(-\left(x-2\right)^2\le0\)

\(\Rightarrow-\left(x-2\right)^2-3\le-3\)

Vậy GTLN của \(B=-3\)

Khi \(x-2=0\)

          \(x=2\)

a)

\(A=\left(x^2-4x+4\right)-11\)

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

Ta có

\(\left(x-2\right)^2-11\ge-11\)

Dấu " = " xảy ra khi x = 2

Vậy MIN_A= - 11 khi x=2

b) 

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

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

Ta có

\(-\left(x-2\right)^2-3\le-3\) với mọi x

Dấu " = " xảy ra khi = 2

Vậy MAX_B= - 3 khi x = 2

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

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

                                       \(=-\left(x^2+2.2x+2^2\right)-3\)

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

             Vì \(-\left(x+2\right)^2\le0\)

Suy ra:\(-\left(x+2\right)^2-3\le-3\)

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

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

                                           \(=-\left(x^2-2.2x+2^2\right)-3\)

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

                  Dấu = xảy ra khi x-2=0

                                              x=2

Vậy MinA=-3 khi x=2

A = x² - 4x + 7 

    = x( x - 4 ) + 7

Vì x( x - 4 ) \(\le\)0

=> Để x( x - 4 ) + 7 \(\le\)7

    => A        \(\ge\)- 7

Vậy GTNN A = - 7 khi x( x - 4 ) = - 7 

Ta có : A = x² - 4x + 7 

= x² - 4x + 4 + 3

A = (x - 2)² + 3 

Vì : \(\left(x-2\right)^2\ge0\forall x\) 

Nên :  A = (x - 2)² + 3   \(\ge3\forall x\)

Vậy A_min = 3 khi x = 2

A=X^2-4x+7

=x^2-2.2x+4+3

=(x-2)^2+3

vì (x-2)^2 lớn hơn hoặc bằng 0 

=> (x-2)^2 +3 lớn hơn hoặc bằng 3

-Dấu "=" xảy ra khi

(x-2)^2=0=> x-2=0=>x=2

Vậy GTNN A=3 khi x=2

B=X^2+8x

=x^2+2.4x+16-16

=(x+4)^2-16

vì (x+4)^2 lớn hơn hoặc bằng 0

=>(x+4)^2-16 sẻ lớn hơn hoặc bằng 16

Dấu "=" xảy ra khi

(x+4)^2=0=>x+4=0=>x=-4

Vậy GTNN B=-16 khi x=-4

C:ca1cl làm cug tương tự như v nha bạn

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

\(B=\left[\left(2x\right)^2-2.2x.1+1^2\right]+6\)

\(B=\left(2x-1\right)^2+6\)

Ta có : \(\left(2x-1\right)^2\ge0\)

\(\Rightarrow\left(2x-1\right)^2+6\ge6\)

Vậy GTNN của B là \(6\)

Khi \(2x-1=0\)

       \(2x=1\)

       \(x=\frac{1}{2}\)

C=2x²+4x+5

C=2.(x²+2x+2,5)

C=2.(x²+2x+1)+3

C=2.(x+1)²+3 

       Vì 2.(x+1)²\(\ge\)0

                  Suy ra:2.(x+1)²+3 \(\ge\)3

Dấu = xảy ra khi x+1=0

                            x=-1

Vậy Min C=3 khi x=-1

B= 4x^2-4x+7

=4x²-4x+1+6

=(2x-1)²+6\(\ge\)6

Dấu = khi x=1/2

Vậy Bmin=6 khi x=1/2

C=2x^2+4x+5

=2x²+4x+2+3

=2(x²+2x+1)+3

=2(x+1)²+3\(\ge\)3

Dấu = khi x=-1

Vậy MinD=3 khi x=-1

I zì:vv

a) Ta có: \(A=4x^2+4x+11=4x^2+4x+1=10=\left(2x+1\right)^2+10\ge10\forall x\)

Vậy Min_A=10 khi \(x=-\dfrac{1}{2}\)

b) Ta có: \(B=5-8x-x^2=-\left(x^2+8x-5\right)=-\left(x^2+8x+16-21\right)\)

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

Vậy Max_B=21 khi x=-4