Question

Tìm giá trị nhỏ nhất của: 

a) A = x² + 4x - 4

b) B = x² + 2y² - 6x + y + 2022

Answer 1

`a)A=x^2+4x-4=x^2+4x+4-8=(x+2)^2-8`

Vì `(x+2)^2 >= 0 AA x<=>(x+2)^2-8 >= -8 AA x`

    Hay `A >= -8 AA x`

 `=>Min_A =-8<=>(x+2)^2=0<=>x=-2`

`b)B=x^2+2y^2-6x+y+2022`

`<=>B=(x^2-6x+9)+2(y^2+1/2y+1/16)+16103/8`

`<=>B=(x-3)^2+2(y+1/4)^2+16103/8`

 Vì `(x-3)^2 >= 0 AA x` ,  `2(y+1/4)^2 >= 0 AA y`

 `<=>(x-3)^2+2(y+1/4)^2 >= 0 AA x,y`

 `<=>(x-3)^2+2(y+1/4)^2+16103/8 >= 16103/8 AA x,y`

   Hay `B >= 16103/8 AA x,y`

    `=>Min_B=16103/8<=>{((x-3)^2=0),((y+1/4)^2=0):}<=>{(x=3),(y=-1/4):}`

Answer 2

a: \(A=x^2+4x+4-8=\left(x+2\right)^2-8>=-8\)

Dấu '=' xảy ra khi x=-2

b: \(B=x^2-6x+9+2\left(y^2-y+\dfrac{1}{4}\right)+\dfrac{4025}{2}\)

\(=\left(x-3\right)^2+2\left(y-\dfrac{1}{2}\right)^2+\dfrac{4025}{2}>=\dfrac{4025}{2}\)

Dấu '=' xảy ra khi x=3 và y=1/2