`a)`

`A=x^2-20x+101`

   `=(x^2-2.x.10+100)+1`

   `=(x-10)^2+1`

Vì `(x-10)^2≥0`  `∀x`

`⇒A≥1`  

Dấu " `=` " xảy ra khi:

        `(x-10)^2=0`

`⇒x-10=0`

`⇒x=10`

Vậy $MIN_A$ `=1` khi `x=10`

`b)`

Đặt: `A=x^2+5y^2-2xy+6x-14y+13`

`= (x^2-2xy+y^2)+(6x-6y)+9 + (4y^2 - 8y+4)`

`= [(x-y)^2 + 2.(x-y).3 + 3^2]+ 4(y^2-2y+1)`

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

Vì `(x-y+3)^2≥0`  `∀x,y`

     `(y-1)^2≥0`  `∀y`

`⇒(x-y+3)^2 + 4(y-1)^2≥0`  `∀x,y`

`⇒A≥0`  `∀x,y`

Mà `A≤0`

Dấu " `=` " xảy ra khi: `{( (x-y+3)^2=0 ),( (y-1)^2=0 ):}`

`⇔{( x-y+3=0 ),( y-1=0 ):}`

`⇔{( x=y-3 ),( y=1 ):}`

`⇔{( x=1-3=-2 ),( y=1 ):}`

Thay `(x;y)=(-2;1)` vào `M,` ta có:

 `M=(-2+3)^2024 + (1-2)^2023`

      `= 1^2024 + (-1)^2023`

       `=1+(-1)`

       `=0`

Vậy `M=0` khi `(x;y)=(-2;1)`

`#` `ccUh`