$Có$ $A=2x^{2}+y^{2}-2x-6y+2xy+33$ 
$=(x^{2}+y^{2}-6x-6y+2xy+9)+(x^{2}+4x+4)+20$
$=(x+y-9)^{2}+(x+2)^{2}+20$ 

$Vì$ $(x+y-9)^{2}$ $≥0∀x$ $và (x+2)^{2} ≥0∀x$

$⇒(x+y-9)^{2}$$+(x+2)^{2}≥0∀x$

$⇒(x+y-9)^{2}$$+(x+2)^{2}+20≥20∀x$

$hay$ $A ≥20$

$A=20^{}⇔$ $\left \{ {{x+y-9=0} \atop {x+2=0}} \right.$ 

$⇔\left \{ {{x+y=9} \atop {x=-2}} \right.$

$⇔\left \{ {{x=-2} \atop {y=11}} \right.$

              $Vậy^{}$  $MinA=20^{}⇔$ $\left \{ {{x=-2} \atop {y=11}} \right.$

$Có$ $B=5x^{2}+y^{2}-8x+4y-2xy-6$ 

$=(x^{2}+y^{2}-4x+4y-2xy+4)+(4x^{2}-4x+1)-11$

$=(x-y-2)^2+(2x-1)^{2}-11$

$Vì$ $(x-y-2)^{2}$ $≥0∀x$ $và (2x-1)^{2} ≥0∀x$

$⇒(x-y-2)^{2}$$+(2x-1)^{2}$  $≥0∀x$

$⇒(x-y-2)^{2}$$+(2x-1)^{2}-11$  $≥-11∀x$

$hay$ $B ≥-11$

$B=-11⇔$ $\left \{ {{x-y-2=0} \atop {2x-1=0}} \right.$ 

              $⇔\left \{ {{x-y=2} \atop {2x=1}} \right.$ 

              $⇔\left \{ {{x=\frac{1}{2}} \atop {y=\frac{-3}{2}}} \right.$ 

        $Vậy^{}$  $MinB$=-11⇔$ $\left \{ {{x=\frac{1}{2}} \atop {y=\frac{-3}{2}}} \right.$

           

 

Đáp án:

Ta có : 

Đề phỉa là $A= 2x ²+y ² -6x-2y+2xy+33$ chớ bạn 

$A = 2x^2 + y^2 - 6x - 2y + 2xy + 33$

$ = [(x^2 + 2xy + y^2) - 2.(x + y) + 1 ] + ( x^2 - 4x + 4) + 28$

$ = [(x + y)^2 - 2(x + y) + 1] + ( x - 2)^2 + 28$

$ = ( x + y - 1)^2 + ( x - 2)^2 + 28$

Do $( x + y - 1)^2 ≥ 0$

     $ ( x - 2) ≥ 0$

$ => ( x + y - 1)^2 + ( x - 2)^2 + 28 ≥ 28$

Dấu "=" xẩy ra

<=> $\left \{ {{ x + y - 1 = 0} \atop {x - 2 = 0}} \right.$ 

<=> $\left \{ {{x + y = 1} \atop {x=2}} \right.$ 

<=> $\left \{ {{y=-1} \atop {x=2}} \right.$ 

Vậy Min A là 28 <=> $\left \{ {{y=-1} \atop {x=2}} \right.$ 

b, Ta có : 

$ B = 5x^2 + y^2 - 8x + 4y - 2xy - 6$

$ = [(x^2 - 2xy + y^2) -4.(x - y) + 4] + (4x^2 - 4x + 1) - 11$

$ = [(x - y)^2 - 4(x - y) + 4] + (2x - 1)^2 - 11$

$ = ( x - y + 2)^2 + (2x - 1)^2 - 11$

Do $( x - y + 2)^2 ≥ 0$

     $ ( 2x - 1)^2 ≥ 0$

$ => ( x - y + 2)^2 + (2x - 1)^2 - 11 ≥ -11$

Dấu "=" xẩy ra

<=> $\left \{ {{ x - y + 2 = 0} \atop {2x - 1 = 0}} \right.$ 

<=> $\left \{ {{x - y = -2} \atop {x= \dfrac{1}{2} }} \right.$

  <=> $\left \{ {{y= \dfrac{5}{2} } \atop {x= \dfrac{1}{2}}} \right.$ 

Vậy Min B là -11 <=> <=> $\left \{ {{y= \dfrac{5}{2} } \atop {x= \dfrac{1}{2}}} \right.$ 

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