\(A=3x^2+6x+7\)

\(A=\left(3x^2+6x+3\right)+4\)

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

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

\(\left(x+1\right)^2\ge0\Rightarrow3\left(x+1\right)^2\ge0\Rightarrow3\left(x+1\right)^2+4\ge4\)

\(\Rightarrow A\ge4\)

dấu "=" xảy ra khi : 

(x + 1)² = 0 => x + 1 = 0 => x = -1

vậy gtnn của A = 4 khi x = -1

`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 = \(\dfrac{-3}{10}\)

Min_A=4 tại x=1

P/s ko chắc.

ta có: A = 3x² - 6x +7

<=> A = 3 ( x² - 2x + 1) +4

<=> A = 3( x-1)² +4

Vì 3( x-1)² >= 0 => 3( x-1)² +4 >= 4

=> Dấu bằng xảy ra <=> 3(x-1)²= 0

<=> x =1

Vậy GTNN của A là 4 khi x = 1

3x²-6x+7

=3x²-6x+3+4

=3(x²-2x+1)+4

=3(x-1)²+4

Min A=4 tại x=1

toi ko bt

A= -4 - x^2 +6x

  =-(x²-6x+9)+5

=-(x-3)²+5\(\le\)5

Dấu "=" xảy ra khi x=3

Vậy...............

B= 3x^2 -5x +7

\(=3\left(x^2-2.\frac{5}{6}x+\frac{25}{36}\right)-\frac{59}{12}\)

\(=3\left(x-\frac{5}{6}\right)^2-\frac{59}{12}\ge\frac{-59}{12}\)

Dấu "=" xảy ra khi \(x=\frac{5}{6}\)

Vậy.................

\(A=x^2+6x+10\)
\(=\left(x^2+2.x.3+3^2\right)-3^2+10\)
\(=\left(x+3\right)^2+1\)
\(Có:\left(x+3\right)^2\ge0\) \(\text{với mọi x}\)
\(\Rightarrow\left(x+3\right)^2+1\ge0+1=1\text{với mọi x}\)
\(\text{GTNN của biểu thức A là 1}\)
\(\text{khi x+3=0 hay x=-3}\)
\(B=3x^2+15x+7\)
\(=3\left(x^2+5x+\frac{7}{3}\right)\)
\(=3\left[x^2+2.x.\frac{5}{2}+\left(\frac{5}{2}\right)^2\right]-\left(\frac{5}{2}\right)^2+\frac{7}{3}\)
\(=3\left(x+\frac{5}{2}\right)^2-\frac{47}{12}\)
\(Có:\left(x+\frac{5}{2}\right)^2\ge0\) \(\text{với mọi x}\)
\(\Rightarrow3\left(x+\frac{5}{2}\right)^2-\frac{47}{12}\ge3.0-\frac{47}{12}=-\frac{47}{12}\text{với mọi x}\)
\(\Rightarrow\text{GTNN của biểu thức B là -}\frac{47}{12}\)
\(\text{khi}x+\frac{5}{2}=0hayx=-\frac{5}{2}\)


 

đặt x^2-7x=y=> \(y\ge-\frac{49}{4}\) (*)

\(A=y\left(y+12\right)=y^2+12y=\left(y+6\right)^2-36\ge-36\)

đẳng thức khi y=-6 thủa mãn đk (*)

Vậy: GTNN của A=-36 khí y=-6 =>\(\left[\begin{matrix}x=1\\x=6\end{matrix}\right.\)

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

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

\(=\left(2x+1\right)^2+\left(y+1\right)^2+5\ge5\)

Dấu = xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=-1\end{matrix}\right.\)

Vậy \(Min_A=5\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=-1\end{matrix}\right.\)

\(B=6x+3x^2+4\)

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

\(=3\left(x+1\right)^2+1\ge1\)

Dấu = xảy ra \(\Leftrightarrow x=-1\)

Vậy \(Min_B=1\Leftrightarrow x=-1\)

\(A=x^2-6x+3\)

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

\(=\left(x+3\right)^2-6\)

ma \(\left(x+3\right)^2\ge0\Leftrightarrow\left(x+3\right)^2-6\ge-6\)

vậy gtnn của A là -6 tại x=-3

\(B=x^2+3x+7=\left(x^2+2.\frac{3}{2}x+\frac{9}{4}\right)+\frac{17}{4}\)

\(=\left(x+\frac{3}{2}\right)^2+\frac{17}{4}\ge\frac{17}{4}\)

vay .............................................

2/

\(A=-x^2+4x+8=-\left(x^2-4x+4\right)+12=-\left(x-2\right)^2+12\le12\)

vay .........................................

\(B=-x^2+3x-5=-\left(x^2-2\frac{3}{2}x+\frac{9}{4}\right)-\frac{11}{4}=\left(x-\frac{3}{2}\right)^2-\frac{11}{4}\le-\frac{11}{4}\)

vay.....................................

nếu có sai mong bạn thông cảm

ko sao cảm ơn

1/ Ta có: A\(=x^2-6x+3\)

\(=x^2-2.x.3+3^2-6\)

\(=\left(x-3\right)^2-6\ge-6\left(\forall x\right)\)

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

Vậy Min A  = -6 khi x = 3.

Ta có: B = \(x^2+3x+7\)

\(=x^2-2.x.\frac{3}{2}+\frac{9}{4}+\frac{19}{4}\)

\(=\left(x-\frac{3}{2}\right)^2+\frac{19}{4}\ge\frac{19}{4}\forall x\)

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

Vậy Min B = 19/4 khi x = 3/2.

2/ 

Ta có: \(A=-x^2+4x+8\)

\(=-\left(x^2-4x-8\right)=-\left(x^2-4x+4-12\right)\)

\(=-\left[\left(x-2\right)^2-12\right]\)

\(=-\left(x-2\right)^2+12\le12\left(\forall x\right)\)

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

Vậy Max A = 12 khi x =2.

Ta có: \(B=-x^2+3x-5\)

\(=-\left(x^2-3x+5\right)=-\left(x^2-3x+\frac{9}{4}+\frac{11}{4}\right)\)

\(=-\left[\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\right]=-\left(x-\frac{3}{2}\right)^2-\frac{11}{4}\le-\frac{11}{4}\left(\forall x\right)\)

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

Vậy Min B = -11/4 khi x =3/2.

Chúc bn hc tốt!

1/

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

Dấu "=" xảy ra khi x=1/2

Vậy Amin=4 khi x=1/2

b, \(B=3x^2+6x-1=3\left(x^2+2x+1\right)-4=3\left(x+1\right)^2-4\ge-4\)

Dấu "=" xảy ra khi x=-1

Vậy Bmin = -4 khi x=-1

2/

a, \(A=10+6x-x^2=-\left(x^2-6x+9\right)+19=-\left(x-3\right)^2+19\le19\)

Dấu "=" xảy ra khi x=3

Vậy Amax = 19 khi x=3

b, \(B=7-5x-2x^2=-2\left(x^2-\frac{5}{2}x+\frac{25}{16}\right)+\frac{31}{8}=-2\left(x-\frac{5}{4}\right)^2+\frac{31}{8}\le\frac{31}{8}\)

Dấu "=" xảy ra khi x=5/4

Vậy Bmax = 31/8 khi x=5/4