\(M=2x^2-16x+32+9y^2-12y+4+1981\)

\(=2\left(x-4\right)^2+\left(3y-2\right)^2+1981\ge1981\forall x,y\)

Dấu '=' xảy ra khi x=4 và y=2/3

Nỗi hứng lm cho vui!

Bài 1:

a) H = \(x^2-4x+16=\left(x^2-4x+4\right)+12=\left(x-2\right)^2+12\)

Vì \(\left(x-2\right)^2\ge0\) => H \(\ge\) 12

=> Dấu = xảy ra <=> \(x=2\)

b) K = \(2x^2+9y^2-6xy-8x-12y+2018\)

= \(\left(x^2-6xy+9y^2\right)+4\left(x-3y\right)+\left(x^2-12x+36\right)+1982\)

= \(\left(x-3y\right)^2+4\left(x-3y\right)+4+\left(x-6\right)^2+1978\)

= \(\left(x-3y+2\right)^2+\left(x-2\right)^2+1978\)

Vì \(\left\{{}\begin{matrix}\left(x-3y+2\right)^2\ge0\\\left(x-6\right)^2\ge0\end{matrix}\right.\) => K \(\ge\) 1978

=> Dấu = xảy ra <=> \(\left\{{}\begin{matrix}y=\dfrac{2+x}{3}\\x=6\end{matrix}\right.\) => \(x=6;y=\dfrac{8}{3}\)

Bài 2:

a) P = \(-x^2-4x+16=-\left(x^2+4x+4\right)+20\)

= \(-\left(x+2\right)^2+20\le20\)

=> Dấu = xảy ra <=> \(x=-2\)

b) \(Q=-x^2+2xy-4y^2+2x+10y-2017\)

= \(-\left[\left(x^2-2xy+y^2\right)+3\left(y^2-4y+4\right)-2\left(x-y\right)+2005\right]\)

= \(-\left[\left(x-y\right)^2-2\left(x-y\right)+1+3\left(y-2\right)^2+2004\right]\)

= \(-\left[\left(x-y-1\right)^2+3\left(y-2\right)^2\right]-2004\)

Vì \(\left\{{}\begin{matrix}-\left(x-y-1\right)^2\le0\\3\left(y-2\right)^2\le0\end{matrix}\right.\) => Q \(\le-2004\)

=> Dấu = xảy ra <=> \(\left\{{}\begin{matrix}x=y+1\\y=2\end{matrix}\right.\) <=> \(x=3;y=2\)

đăng từng câu 1 thôi

B = x²y²+2x²+24xy+16x+191 = [ (xy)^2 + 24xy + 144] + \(\left[\left(\sqrt{2}x\right)^2+2.\sqrt{2}x.4\sqrt{2}+32\right]\)+15

= (xy+12)^2 +(\(\sqrt{2}x\)+\(4\sqrt{2}\))^2 + 15 

( ở đây mik làm tắt) => Min B = 15 khi \(\sqrt{2}x+4\sqrt{2}=0=>x=-4\)và xy+12 = 0 => -4y = -12= > y=3

A= 2x^2+9y^2-6xy-6x-12y+2004

A = (x^2 -6xy +9y^2) + 4(x -3y) + x^2 - 10x + 2004

A = [(x -3y)^2 +4(x -3y) + 4] + (x^2 -10x +25) + 1975

A= (x -3y +2)^2 + (x -5)^2 + 1975

( mik rút mấy cái bước (x-3y+2)^2 = 0, bn làm thì nên thêm vào=> Min A = 1975 vs x= 5 và y = 7/3

D=-x^2+2xy-4y^2+2x+10y-8

D = (-x^2 - y^2 - 1 + 2xy + 2x - 2y) + (-3y^2 + 12y - 12) + 5

D = -(x^2+y^2+1 - 2xy - 2x + 2y) - 3(y^2 - 4y + 4) + 5

D= - (x - y - 1)^2 - 3(y - 2)^2 +5 

=> Max D = 5 khi x= 3 và y=2

C = x^2 + y^2 + 9z^2 - 2x + 12y + 6z + 24

C = (x^2-2x+1)+(y^2 + 2y.6 + 36)+ [ (3z)^2 + 2.3z + 1] - 14

C= (x-1)^2 + (y+6)^2 + (3z+1)^2 - 14

....

mà chép lại đề câu B cho mik vs

d)  D = x⁴ - 6x² + 10

D = (X²)² - 2. x². 3 + 3² + 1

D = (x² - 3)² + 1

(x² - 3)²  >= 0 với mọi x

(x² - 3)² + 1 >=1 với moi5 x

Vậy GTNN của D là 1

câu A thiếu đề

B=\(x^2-2x+2017=\left(x-1\right)^2+2016>=2016\)

Min B=2016 khi x-1=0<=>x=1

+)D=\(-2x^2+4x+2017=-2\left(x^2-2x+1\right)+2019=-2\left(x-1\right)^2+2019< =2019\)

=>Max D=2019, dấu '=' xảy ra khi x-1=0<=>x=1

Bổ sung câu A. \(A=x^2+2xy+3y^2-4y+2017\)

\(A=x^2+2xy+3y^2-4y+2017\)

\(A=\left(x^2+2xy+y^2\right)+\left(2y^2-4y+2\right)+2015\)

\(A=\left(x^2+2xy+y^2\right)+2\left(y^2-2y+1\right)+2015\)

\(A=\left(x+y\right)^2+2\left(y-1\right)^2+2015\ge2015\)

Vậy A_min=2015 <=> x=-1 và y=1

a) Ta có: \(Q=-x^2-y^2+4x-4y+2=-\left(x^2+y^2-4x+4y-2\right)\)

\(=-\left(x^2-4x+4+y^2+4y+4\right)+10\)

\(=-\left[\left(x-2\right)^2+\left(y+2\right)^2\right]+10\le10\forall x,y\)

Vậy Max_Q=10 khi x=2, y=-2

b) +Ta có: \(A=-x^2-6x+5=-\left(x^2+6x-5\right)=-\left(x^2+6x+9-14\right)\)

\(=-\left(x^2+6x+9\right)+14=-\left(x+3\right)^2+14\le14\forall x\)

Vậy Max_A=14 khi x=-3

+Ta có: \(B=-4x^2-9y^2-4x+6y+3=-\left(4x^2+9y^2+4x-6y-3\right)\)

\(=-\left(4x^2+4x+1+9y^2-6y+1-5\right)\)

\(=-\left[\left(2x+1\right)^2+\left(3y-1\right)^2\right]+5\le5\forall x,y\)

Vậy Max_B=5 khi x=-1/2, y=1/3

c) Ta có: \(P=x^2+y^2-2x+6y+12=x^2-2x+1+y^2+6y+9+2\)

\(=\left(x-1\right)^2+\left(y+3\right)^2+2\ge2\forall x,y\)

Vậy Min_P=2 khi x=1, y=-3