Bài 4:

a,

`A=(x-2y)^2+(y-2021)^2020`

Vì `(x-2y)^2>=0` với mọi `x`

`(y-2021)^2020>=0` với mọi `y`

`=>A=(x-2y)^2+(y-2021)^2020>=0`

Dấu `=` xảy ra `<=>`$\left \{ {{x-2y=0} \atop {y-2021=0}} \right.$`<=>` $\left \{ {{x=2y} \atop {y=2021}} \right.$`<=>` $\left \{ {{x=4042} \atop {y=2021}} \right.$ 

Vậy `A_(min)=0<=>x=4042; y=2021`

b,

`B=(x+y-3)^4+(x-2y)^2+2021>=2021`

Dấu `=` xảy ra `<=>`$\left \{ {{x+y-3=0} \atop {x-2y=0}} \right.$`<=>` $\left \{ {{x=3-y} \atop {x=2y}} \right.$ `<=>`$\left \{ {{x=3-y} \atop {3-y=2y}} \right.$`<=>` $\left \{ {{x=3-y} \atop {y=1}} \right.$ `<=>`$\left \{ {{x=2} \atop {y=1}} \right.$ 

Vậy `B_(min)=2021<=>x=2; y=1`

Bài 1:

a,

`A=(x-2)^2+4>=4`

Dấu `=` xảy ra `<=>x-2=0<=>x=2`

Vậy `A_(min)=4<=>x=2`

b,

`B=-5+x^4+x^2>=-5`

Dấu `=` xảy ra `<=>x^4=x^2=0<=>x=0`

c,

`C=3|x-1|-12>=-12`

Dấu `=` xảy ra `<=>x-1=0<=>x=1`

Vậy `C_(min)=-12<=>x=1`

d,

`D=2|x-2|+(y-3)^2+10>=10`

Dấu `=` xảy ra `<=>`$\left \{ {{x-2=0} \atop {y-3=0}} \right.$`<=>` $\left \{ {{x=2} \atop {y=3}} \right.$ 

Vậy `D_(min)=10<=>x=2; y=3`

Bài 4:

a,

A=(x−2y)2+(y−2021)2020A=(x-2y)2+(y-2021)2020

Vì (x−2y)2≥0(x-2y)2≥0 với mọi xx

(y−2021)2020≥0(y-2021)2020≥0 với mọi yy

⇒A=(x−2y)2+(y−2021)2020≥0⇒A=(x-2y)2+(y-2021)2020≥0

Dấu == xảy ra ⇔⇔{x−2y=0y−2021=0{x−2y=0y−2021=0⇔⇔ {x=2yy=2021{x=2yy=2021⇔⇔ {x=4042y=2021{x=4042y=2021 

Vậy Amin=0⇔x=4042;y=2021Amin=0⇔x=4042;y=2021

b,

B=(x+y−3)4+(x−2y)2+2021≥2021B=(x+y-3)4+(x-2y)2+2021≥2021

Dấu == xảy ra ⇔⇔{x+y−3=0x−2y=0{x+y−3=0x−2y=0⇔⇔ {x=3−yx=2y{x=3−yx=2y ⇔⇔{x=3−y3−y=2y{x=3−y3−y=2y⇔⇔ {x=3−yy=1{x=3−yy=1 ⇔⇔{x=2y=1{x=2y=1 

Vậy Bmin=2021⇔x=2;y=1Bmin=2021⇔x=2;y=1

Bài 1:

a,

A=(x−2)2+4≥4A=(x-2)2+4≥4

Dấu == xảy ra ⇔x−2=0⇔x=2⇔x-2=0⇔x=2

Vậy Amin=4⇔x=2Amin=4⇔x=2

b,

B=−5+x4+x2≥−5B=-5+x4+x2≥-5

Dấu == xảy ra ⇔x4=x2=0⇔x=0⇔x4=x2=0⇔x=0

c,

C=3|x−1|−12≥−12C=3|x-1|-12≥-12

Dấu == xảy ra ⇔x−1=0⇔x=1⇔x-1=0⇔x=1

Vậy Cmin=−12⇔x=1Cmin=-12⇔x=1

d,

D=2|x−2|+(y−3)2+10≥10D=2|x-2|+(y-3)2+10≥10

Dấu == xảy ra ⇔⇔{x−2=0y−3=0{x−2=0y−3=0⇔⇔ {x=2y=3{x=2y=3 

Vậy Dmin=10⇔x=2;y=3Dmin=10⇔x=2;y=3