tìm Min A= 2x ²+y ² -2x-6y+2xy+33 B= 5x ²+y ² -8x+4y-2xy-6

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

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

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

              

    

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

   Giải thích các bước giải:

    

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