\(A=x^2+4x+100\)

\(A=x^2+2.x.2+2^2+96\)

\(A=\left(x+2\right)^2+96\)

           \(\left(x+2\right)^2+96\le0\)

           \(\left(x+2\right)^2+96\le96\)

    \(\Leftrightarrow A\le96\)

\(A_{min}\Leftrightarrow A=10\)

Dấu "=" xảy ra : \(\left(x+2\right)^20\)

                             \(x+2=0\)

                             \(x=-2\)

Thay hộ mik cái dấu \(\le\)thành dấu \(\ge\)vs ak

1,A=(x²-6x+9)+2

=(x-3)²+2

ta thấy (x-3)²>=0 với mọi x

=>(x-3)²+2>=2 với mọi x

hay A>=2

dấu "="xảy ra x-3=0<=>x=3

vậy MinA=2 khi x=3

ý b sai đầu bài bạn nhé

C=-(x²-5x)

=-(x²-5x+25/4)+25/4

=-(x-5/2)²+25/4

ta thấy -(x-5/2)²<=0 với mọi x

=>-(x-5/2)²+25/4 <=25/4 với mọi x

hay C<=25/4

dấu "=" xảy ra khi x-5/2=0<=>x=5/2

vậy MaxC=25/4 khi x=5/2

k mk nha

Ta có : A = x² - 6x + 11

<=> A = x² - 6x + 9 + 2 

<=> A = (x - 3)² + 2

Mà (x - 3)² \(\ge0\forall x\)

Nên A =  (x - 3)² + 2 \(\ge2\forall x\)

Vậy A_min = 2 , dấu "=" xảy ra khi và chỉ khi x = 3

a) A = 5x² - 20x + 2020 = 5(x² - 4x + 4) + 2000 = 5(x - 2)² + 2000 \(\ge\)2000 \(\forall\)x

Dấu "=" xảy ra <=> x  - 2 = 0 <=> x = 2

Vậy MinA = 2000 khi x = 2+

b) B = -3x² - 6x + 15 = -3(x² + 2x + 1) + 18 = -3(x + 1)² + 18 \(\le\)18 \(\forall\)x
Dấu "=" xảy ra <=> x + 1 = 0 <=> x = -1

Vậy MaxB = 18 khi x = -1

c) C = 9x² + 2x + 7 = (9x² + 2x + 1/9) + 62/9 = (3x  + 1/3)²  + 62/9 \(\ge\)62/9 \(\forall\)x

Dấu "=" xảy ra <=> 3x + 1/3 = 0 <=> x  = -1/9

Vậy MinC = 62/9 khi x = -1/9

d) D = 16 - 2x² - 8x = -2(x² + 4x + 4) + 24 = -2(x + 2)² + 24 \(\le\) 24 \(\forall\)x

Dấu "=" xảy ra <=> x + 2 = 0 <=> x = -2

Vậy MaxD = 24 khi x = -2

a) \(A=x^2+3x+4=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)

\(minA=\dfrac{7}{4}\Leftrightarrow x=-\dfrac{3}{2}\)

b) \(B=2x^2-x+1=2\left(x-\dfrac{1}{4}\right)^2+\dfrac{7}{8}\ge\dfrac{7}{8}\)

\(minB=\dfrac{7}{8}\Leftrightarrow x=\dfrac{1}{4}\)

c) \(C=5x^2+2x-3=5\left(x+\dfrac{1}{5}\right)^2-\dfrac{16}{5}\ge-\dfrac{16}{5}\)

\(minC=-\dfrac{16}{5}\Leftrightarrow x=-\dfrac{1}{5}\)

d) \(D=4x^2+4x-24=\left(2x+1\right)^2-25\ge-25\)

\(minD=-25\Leftrightarrow x=-\dfrac{1}{2}\)

e) \(E=x^2+6x-11=\left(x+3\right)^2-20\ge-20\)

\(minE=-20\Leftrightarrow x=-3\)

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

\(minG=-\dfrac{4}{3}\Leftrightarrow x=-2\)

\(A=x^2+3x+4=\left(x^2+3x+\dfrac{9}{4}\right)+\dfrac{7}{4}=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\)

Do \(\left(x+\dfrac{3}{2}\right)^2\ge0\forall x\)

\(\Rightarrow A=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)

\(minA=\dfrac{7}{4}\Leftrightarrow x+\dfrac{3}{2}=0\Leftrightarrow x=-\dfrac{3}{2}\)

Mấy câu còn lại làm tương tự nhé em^^

Để A lớn nhất thì tử phải nhỏ nhất hay \(x^2+3x+2\) nhỏ nhất

\(x^2+3x+2=x^2+2\cdot\frac{3}{2}+\frac{9}{4}+2-\frac{9}{4}\)

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

Vì \(\left(x+\frac{3}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+\frac{3}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)

Dấu "=" xảy ra khi\(x+\frac{3}{2}=0\Leftrightarrow x=-\frac{3}{2}\)

Min \(x^2+3x+2=-\frac{1}{4}\) khi x=-3/2

Vậy 

\(MaxA=\frac{2}{-\frac{1}{4}}=2\cdot\left(-4\right)=-8\)

a)4x²-4x+3

=[(2x)²-4x+1]+2

=(2x+1)²+2 \(\ge\)2 với mọi x

Vậy GTNN của 4x²-4x+3 là 2 tại 

(2x+1)²+2=2

<=>(2x+1)²     =0

<=>2x+1       =0

<=>x             =\(\frac{-1}{2}\)

b)-x²+2x-3

=(-x²+2x-1)-2

= -(x²-2x+1)-2

=-(x-1)²-2 \(\le\)-2

Vậy GTLN của -x²+2x-3 là -2 tại :

-(x-1)²-2=-2

<=>-(x-1)²  =0

<=>x-1      =0

<=>x         =1

Bài 1 :

a) \(A=x^2-6x+11\)

\(A=x^2-2\cdot x\cdot3+3^2+2\)

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

Dấu "=' xảy ra \(\Leftrightarrow x-3=0\Leftrightarrow x=3\)

b) \(B=2x^2+10x-1\)

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

\(B=2\left[x^2+2\cdot x\cdot\frac{5}{2}+\left(\frac{5}{2}\right)^2-\frac{27}{4}\right]\)

\(B=2\left[\left(x+\frac{5}{2}\right)^2-\frac{27}{4}\right]\)

\(B=2\left(x+\frac{5}{2}\right)^2-\frac{27}{2}\ge\frac{-27}{2}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x+\frac{5}{2}=0\Leftrightarrow x=\frac{-5}{2}\)

c) \(C=5x-x^2\)

\(C=-\left(x^2-5x\right)\)

\(C=-\left[x^2-2\cdot x\cdot\frac{5}{2}+\left(\frac{5}{2}\right)^2-\left(\frac{5}{2}\right)^2\right]\)

\(C=-\left[\left(x-\frac{5}{2}\right)^2-\frac{25}{4}\right]\)

\(C=\frac{25}{4}-\left(x-\frac{5}{2}\right)^2\le\frac{25}{4}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\)

Bài 2 :

\(\left(x+y+z\right)^3-x^3-y^3-z^3\)

\(=\left[x+\left(y+z\right)\right]^3-x^3-y^3-z^3\)

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

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

\(=3x^2\left(y+z\right)+3x\left(y+z\right)^2+3yz\left(y+z\right)\)

\(=3\left(y+z\right)\left[x^2+x\left(y+z\right)+yz\right]\)

\(=3\left(y+z\right)\left(x^2+xy+xz+yz\right)\)

\(=3\left(y+z\right)\left[x\left(x+y\right)+z\left(x+y\right)\right]\)

\(=3\left(y+z\right)\left(x+y\right)\left(x+z\right)\)

a) A=x²-6x+11

=(x²-6x+9)+2

=(x-3)²+2

Ta có  \(\left(x-3\right)^2\le0vớim\text{ọi}x\)

=>\(\left(x-3\right)^2+2\le2v\text{ới}m\text{ọi}x\)

Dấu "="xảy ra khi : x-3=0

=>x=3

Vậy x có GTNN là 2 tại x=3

a: Ta có: \(A=x^2-2xy+5y^2+4y+51\)

\(=x^2-2xy+y^2+4y^2+4y+1+50\)

\(=\left(x-y\right)^2+\left(2y+1\right)^2+50\ge50\forall x,y\)

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

a) \(A=x^2-2xy+5y^2+4y+51=\left(x^2-2xy+y^2\right)+\left(4y^2+4y+1\right)+50=\left(x-y\right)^2+\left(2y+1\right)^2+50\ge50\)

\(minA=50\Leftrightarrow x=y=-\dfrac{1}{2}\)

c) \(C=\dfrac{9}{-2x^2+4x-7}=\dfrac{9}{-2\left(x^2-2x+1\right)-5}=\dfrac{9}{-2\left(x-1\right)^2-5}\ge\dfrac{9}{-5}=-\dfrac{9}{5}\)

\(minC=-\dfrac{9}{5}\Leftrightarrow x=1\)

d) \(10x^2+4y^2-4xy+8x-4y+20=\left[4y^2-4y\left(x+1\right)+\left(x+1\right)^2\right]+\left(9x^2+6x+1\right)+18=\left(2y-x-1\right)^2+\left(3x+1\right)^2+18\ge18\)

\(minD=18\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-\dfrac{1}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\)

e) \(E=9x^2+2y^2+6xy-6x-8y+10=\left[9x^2+6x\left(y-1\right)+\left(y-1\right)^2\right]+\left(y^2-6x+9\right)=\left(3x+y-1\right)^2+\left(y-3\right)^2\ge0\)

\(minE=0\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-\dfrac{2}{3}\\y=3\end{matrix}\right.\)

\(\dfrac{3x^2-1}{x^2+2}=\dfrac{6x^2-2}{2\left(x^2+2\right)}=\dfrac{7x^2-\left(x^2+2\right)}{2\left(x^2+2\right)}=\dfrac{7x^2}{2\left(x^2+2\right)}-\dfrac{1}{2}\ge=-\dfrac{1}{2}\)

GTNN của biểu thức là \(-\dfrac{1}{2}\), xảy ra khi \(x=0\)

Biểu thức ko tồn tại GTLN