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`a,`

`2x^2 + 2y^2 - 2xy + 2x + 2y+35`

`= (x^2 - 2xy + y^2) + (x^2 + 2x + 1) + (y^2 +2y+1) + 33`

`= (x-y)^2 + (x+1)^2 + (y+1)^2 +33 ≥33∀x,y`

Dấu "`=`" xảy ra khi :

`↔ (x-y)^2=0, (x+1)^2=0, (y+1)^2=0`

`↔x-y=0, x+1=0, y+1=0`

`↔x=y, x=-1,y=-1`

`↔ x=y=-1`

Vậy GTNN của BT là `33 ↔ (x;y)=(-1;-1)`

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`b,`

`(x-1)^2 + (x+2)^2`

`= (x^2 - 2x +1) + (x^2 + 4x + 4)`

`= x^2 - 2x + 1 +x^2 +4x+4`

`= 2x^2 +2x + 5`

`= 2 (x^2 + x + 5/2)`

`= 2 (x^2 + 2 . x . 1/2 + 1/4 + 9/4)`

`= 2 (x+1/2)^2 + 9/2 ≥ 9/2 ∀x`

Dấu "`=`" xảy ra khi :

`↔ (x+1/2)^2=0`

`↔x+1/2=0`

`↔x=(-1)/2`

Vậy GTNN của BT là `9/2 ↔ x=(-1)/2`

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`c,`

`x^2 + y^2 -x+6y+20`

`= (x^2 - x) + (y^2 + 6y)+20`

`= (x^2 - 2 . x . 1/2 + 1/4) + (y^2 + 2 . y . 3 + 9) + 43/4`

`= (x-1/2)^2 + (y+3)^2 + 43/4 ≥ 43/4 ∀x,y`

Dấu "`=`" xảy ra khi :

`↔ (x-1/2)^2=0 ,(y+3)^2=0`

`↔x-1/2=0,y+3=0`

`↔x=1/2,y=-3`

Vậy GTNN của BT là `43/4 ↔ (x;y)=(1/2; -3)`

`#tnvt`

`a)2x^2+2y^2-2xy+2x+2y+35`

`=x^2-2xy+y^2+x^2+2x+1+y^2+2y+1+33`

`=(x-y)^2+(x+1)^2+(y+1)^2+33`

Với mọi `x,y\inRR,` ta có: `{((x-y)^2>=0),((x+1)^2>=0),((y+1)^2>=0):}`

`=>(x-y)^2+(x+1)^2+(y+1)^2+33>=33`

Dấu `=` xảy ra khi `{(x-y=0),(x+1=0),(y+1=0):}`

                              `<=>{(x=y),(x=-1),(y=-1):}`

Vậy `GTNNNN=33` khi `x=y=-1`

`b)(x-1)^2+(x+2)^2`

`=x^2-2x+1+x^2+4x+4`

`=2x^2+2x+5`

`=2.x^2+2.2.x. 1/2 +2.1/4+9/2`

`=2(x^2+2.x. 1/2+1/4)+9/2`

`=2(x+1/2)^2+9/4`

Với mọi `x\inRR,` ta có: `(x+1/2)^2>=0`

`=>2(x+1/2)^2+9/4>=9/4`

Dấu `=` xảy ra khi `x+1/2=0<=>x=-1/2`

Vậy `GTNNNN=9/4` khi `x=-1/2`

`c)x^2+y^2-x+6y+20`

`=x^2-2.x. 1/2+1/4+y^2+2.y.3+9+43/4`

`=(x-1/2)^2+(y+3)^2+43/4`

Với mọi `x,y\inRR,` ta có: `{((x-1/2)^2>=0),((y+3)^2>=0):}`

`=>(x-1/2)^2+(y+3)^2+43/4>=43/4`

Dấu `=` xảy ra khi `x-1/2=0<=>x=1/2`

                              `y+3=0<=>y=-3`

Vậy `GTNNNN=43/4` khi `(x;y)=(1/2;-3)`