\(A=x^2-6x+10\)

\(\Leftrightarrow A=x^2-2\cdot x\cdot3+3^2-9+10\)

\(\Leftrightarrow A=\left(x-3\right)^2+1\ge1\)     \(\forall x\in z\)

\(\Leftrightarrow A_{min}=1khix=3\)

\(B=3x^2-12x+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x\right)^2-2\cdot\sqrt{3}x\cdot2\sqrt{3}+\left(2\sqrt{3}\right)^2-12+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x-2\sqrt{3}\right)^2-11\ge-11\)    \(\forall x\in z\)

\(\Leftrightarrow B_{min}=-11khix=2\)

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

a) x² - 2x + 5 = (x - 1)² + 4 >= 4

Min là 4 khi x = 1

a) \(A=x^2+6x+10\)

\(A=x^2+2\cdot x\cdot3+3^2+1\)

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

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

b) \(B=2x^2+y^2+2xy+4x+15\)

\(B=\left(x^2+2xy+y^2\right)+\left(x^2+2\cdot x\cdot2+2^2\right)+11\)

\(B=\left(x+y\right)^2+\left(x+2\right)^2+11\ge11\forall x;y\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x+y=0\\x+2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}y=2\\x=-2\end{cases}}\)

Bài 1:

a: A=x^2-6x+10

=x^2-6x+9+1

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

Dấu = xảy ra khi x=3

b: \(B=3x^2-12x+1\)

=3(x^2-4x+1/3)

=3(x^2-4x+4-11/3)

=3(x-2)^2-11>=-11

Dấu = xảy ra khi x=2

a. A=x²-6x+13

=x²-2.x.3+3²+4

=(x-3)²+4 > 4

=> A có GTNN là 4

<=> x-3=0

<=> x=3

b. B=4x-x²

=-x²+4x-4+4

=-(x²-4x+4)+4

=-(x-2)²+4 < 4

=> GTLN của B là 4

<=> x-2=0

<=> x=2

****(tick) cho mình nhé!

a)3

b)2

toi ko bt

có ai làm NY tui hem

\(4x^2+4x+6\)

\(=\left(2x\right)^2+2.2x.1+1+5\)

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

\(Min=5\Leftrightarrow2x+1=0\Rightarrow x=\frac{-1}{2}\)

\(x^2+6x+11\)

\(=x^2+2.x.3+9+2\)

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

\(Min=2\Leftrightarrow x+3=0\Rightarrow x-3\)

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

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

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

\(MIn=\frac{-5}{4}\Leftrightarrow x+\frac{3}{2}=0\Rightarrow x=\frac{-3}{2}\)

B = 4x² + 4x - 6 = (2x)² + 2.2.x + 1 - 7 = (2x + 1)² - 7 \(\ge\)-7

             Vậy MinB = -7 khi 2x + 1 = 0 => x = -1/2 

C = x² + 6x + 11 = x² + 2.3.x + 9 + 2 = (x + 3)² + 2 \(\ge\)2

              Vậy MinC = 2 khi x + 3 = 0 => x = -3

D = x² - 3x + 1 \(=x^2-2.\frac{3}{2}.x+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2+1=\left(x-\frac{3}{2}\right)^2-\frac{5}{4}\ge-\frac{5}{4}\)

              Vậy MinD = -5/4 khi x - 3/2 = 0 => x = 3/2

bài a của  o0o I am a studious person o0o có lẽ sai

\(B=4x^2+4x-6=\left(4x^2+4x+1\right)-7=\left(2x+1\right)^2-7\)

có:\(\left(2x+1\right)^2\ge0\)

vậy GTNN của B = -7 tại x = -1/2

có ai giúp em không