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
