mình trả lời ở dưới rồi nha ^^

\(F=x^2-4x+y^2-8y+6\)

\(=\left(x^2-4x+4\right)+\left(y^2-8y+16\right)-14\)

\(=\left(x-2\right)^2+\left(y-4\right)^2-14\)

Nhận xét :

\(\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\\\left(y-4\right)^2\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\)

\(\Leftrightarrow F\ge-14\)

Dấu "=" xảy ra khi \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)

Vậy..

Ta có:

x² - 4x + y² - 8y + 6

= (x² - 2.2x + 2²) + (y² - 2.4y + 4²) - 14

= (x - 2)² + (y - 4)² - 14 > = -14

Vậy F_Min = -14 <=> (x;y) = (2;4)

a) \(A=x^2+6x+11\)

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

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

Có: \(\left(x+3\right)^2\ge0\Rightarrow\left(x+3\right)^2+2\ge2\)

Dấu = xảy ra khi: \(\left(x+3\right)^2=0\Rightarrow x+3=0\Rightarrow x=-3\)

Vậy: \(Min_A=2\) tại \(x=-3\)

b) \(B=4x-x^2+1\)

\(B=-x^2+4x-4+5\)

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

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

Có: \(\left(x-2\right)^2\ge0\)

\(\Rightarrow5-\left(x-2\right)^2\le5\)

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

Vậy: \(Max_B=5\) tại \(x=2\)

d, (x-1) (x+2) (x+3) (x+6)
=(x^2+2x-x-2) (x^2+6x+3x+18)
=(x^2-x^2) + (2x-x+6x-3x) = (-2+18)
=0            + (-8x)              =16
=                    x                =16:(-8)
=                  x                  =-2

A = x^2 + 6x + 11

= x^2 + 6x + 9 + 2

= (x + 3)^2 + 2

min = 2

a, A = x^2 + 6x + 11

= x^2 + 6x + 9 + 2

= (x + 3)^2 + 2

làm tiếp

b, x^2 - 20x + 101

= x^2  20x + 100 + 1

= (x - 10)^2 + 1

có (x - 10)^2 > 0 => (x - 10)^2 +  > 1

+) =\(\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+3=\left(x-1\right)^2+\left(y+2\right)^2+3\ge3\)

=> GTNN =3 khi x=1 và y=-2

+) =\(\left(x^2-4x+4\right)+\left(y^2-8y+16\right)-14=\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\)

=> GTNN =-14 khi x=2 và y=4

e) ta có: \(x^2-2x+y^2+4y+8=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+3\)

                                                      = \(\left(x-1\right)^2+\left(y+2\right)^2+3\)\(\ge3\)

vậy min =3 . dấu = khi x=1; y=-2

f) ta có:\(x^2-4x+y^2-8y+6=\left(x^2-4x+4\right)+\left(y^2-8y+16\right)-14\)                                                                                                                                                      =\(\left(x-2\right)^2+\left(y-4\right)^2+\left(-14\right)\ge\left(-14\right)\)

vậy min =-14 khi x=2;y=4.

a,<=>   x²-4x+2²+y²-8y+4²-14

<=> (x²-2x2+2²)+(y²-2x4+4²)-14

<=> (x-2)²+(y-4)²-14 

Vì (x-2)²+(y-4)²>= 0

=> F >= -14 => MIn F = -14 <=> x=2, y=4

b, <=> (x²+5²+(2y)²-4xy+10x-20y) +(y²-2y+1)+2

<=> (x+5-2y )²+(y-1)²+2 

Vì (x+5-2y) ²+(y-1)² >= 0

=> G >= 2 => Min =2 <=> y=1, x= -3

\(F=x^2-4x+y^2-8y+6\)

\(F=\left(x^2-2.2x+2^2\right)+\left(y^2-2.4.y+4^2\right)-14\)

\(F=\left(x-2\right)^2+\left(y-4\right)^2-14\)

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

\(\left(y-4\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\forall x\)

\(F=-14\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^2=0\\\left(y-4\right)^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=4\end{cases}}\)

Vậy \(F_{min}=-14\Leftrightarrow\hept{\begin{cases}x=2\\y=4\end{cases}}\)

Ta có:

A = (x + 2)² + (x - 3)² = x² + 4x + 4 + x² - 6x + 9 = 2x² - 2x + 13 = 2(x² - x + 1/4) + 25/2 = 2(x - 1/2)² + 25/2

Ta luôn có: (x - 1/2)² \(\ge\) 0 \(\forall\)x  ----> 2(x - 1/2)² \(\ge\) 0 \(\forall\)x

  => 2(x - 1/2)² + 25/2 \(\ge\) 25/2 \(\forall\)x

Dấu "=" xảy ra khi: (x - 1/2)² = 0 <=> x - 1/2 = 0 <=> x = 1/2

Vậy A_min = 25/2 tại x = 1/2

B = x² - 4x + y² - 8y + 6 = (x² - 4x + 4) + (y² - 8y + 16) - 14 = (x - 2)² + (y - 4)²  - 14

Ta luôn có: (x - 2)² \(\ge\)0 \(\forall\)x

                 (y - 4)² \(\ge\)0 \(\forall\)y

=> (x - 2)² + (y - 4)² - 14 \(\ge\) -14 \(\forall\)x,y

hay B \(\ge\)-14 \(\forall\)x, y

Dấu "=" xảy ra khi : \(\hept{\begin{cases}\left(x-2\right)^2=0\\\left(y-4\right)^2=0\end{cases}}\) <=> \(\hept{\begin{cases}x-2=0\\x-4=0\end{cases}}\) <=> \(\hept{\begin{cases}x=2\\y=4\end{cases}}\)

Vậy B_min = -14 tại x = 2 và y = 4

a)Vì \(\left(x+2\right)^2\ge0\)

nên \(\left(x+2\right)^2\)nhỏ nhất khi \(x=-2\)

thay vào A ta có: \(\left(-2+2\right)^2+\left(-2-3\right)^2=\left(-5\right)^2=25\left(1\right)\)

Vì \(\left(x-3\right)^2\ge0\)

nên \(\left(x-3\right)^2\)nhỏ nhất khi \(x=3\)

Thay vào A ta có: \(\left(3+2\right)^2+\left(3-3\right)^2=5^2=25\left(2\right)\)

So sánh (1) và (2) ta có: (1) = (2)

Vậy \(GTNN_A=25\)tại \(x=-2\)hay \(x=3\)

b)\(B=x^2-4x+y^2-8y+6\)

\(B=x^2-4x+y^2-8y+4+16-14\)

\(B=\left(x^2-4x+4\right)+\left(y^2-8y+16\right)-14\)

\(B=\left(x-2\right)^2+\left(y-4\right)^2-14\)

Vì \(\left(x-2\right)^2\ge0\)

\(\left(y-4\right)^2\ge0\)

\(\Rightarrow B=\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\)

Vậy \(GTNN_B=-14\)tại \(x=2\)và \(y=4\)

c) \(C=x^2-2x+y^2+4y+8\)

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

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

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

Vì \(\left(x-1\right)^2\ge0\)

\(\left(y+2\right)^2\ge0\)

\(\Rightarrow C=\left(x-1\right)^2+\left(y+2\right)^2+3\ge3\)

Vậy \(GTNN_C=3\)tại \(x=1\)và \(y=-2\)