\(T=-2\left(x^2+y^2+1-2xy+2x-2y\right)-2y^2+8y+2004\)

\(T=-2\left(x-y+1\right)^2-2\left(y-2\right)^2+2012\le2012\)

\(T_{max}=2012\) khi \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

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nhìn đề bài rắc rối thế

A=x2−2xy−4x+2y2−8y+20C=x2−2xy−4x+2y2−8y+20

=(x2−2xy+y2)−4(x−y)+4+y2−12y+36−20=(x2−2xy+y2)−4(x−y)+4+y2−12y+36−20

=(x−y)2−4(x−y)+4+(y−6)2−20=(x−y)2−4(x−y)+4+(y−6)2−20

=(x−y−2)2+(y−6)2−20≥−20∀x;y=(x−y−2)2+(y−6)2−20≥−20∀x;y

Dấu " = " xảy ra ⇔{x−y−2=0y−6=0⇔{x−y=2y=6⇔{x−y−2=0y−6=0⇔{x−y=2y=6

⇔x=8;y=6⇔x=8;y=6

Vậy  A là : −20⇔x=8;y=6

Bài 1

a) \(A=\left(x+1\right)\left(2x-1\right)=2x^2+x-1=2\left(x^2+\frac{x}{2}-\frac{1}{2}\right)=2\left(x^2+2.\frac{1}{4}.x+\frac{1}{16}-\frac{9}{16}\right)\)\(=2\left[\left(x+\frac{1}{4}\right)^2-\frac{9}{16}\right]=2\left(x+\frac{1}{4}\right)^2-\frac{9}{8}\)

Vì \(\left(x+\frac{1}{4}\right)^2\ge0\Rightarrow2\left(x+\frac{1}{4}\right)^2\ge0\Rightarrow2\left(x+\frac{1}{4}\right)^2-\frac{9}{8}\ge-\frac{9}{8}\)

Dấu "=" xảy ra khi \(\left(x+\frac{1}{4}\right)^2=0\Leftrightarrow x+\frac{1}{4}=0\Leftrightarrow x=-\frac{1}{4}\)

Vậy minA=-9/8 khi x=-1/4

b)\(B=4x^2-4xy+2y^2+1=\left(4x^2-4xy+y^2\right)+y^2+1=\left(2x-y\right)^2+y^2+1\)

Vì \(\hept{\begin{cases}\left(2x-y\right)^2\ge0\\y^2\ge0\end{cases}}\)=>\(\left(2x-y\right)^2+y^2\ge0\Rightarrow B=\left(2x-y\right)^2+y^2+1\ge1\)

Dấu "=" xảy ra khi (2x-y)²=y²=0 <=> 2x-y=y=0 <=> x=y=0

Vậy minB=1 khi x=y=0

lý luận tương tự bài 1, bài này mình làm tắt

Bài 2:

a) \(C=5x-3x^2+2=-\left(3x^2-5x-2\right)=-3\left(x^2-\frac{5}{3}x-\frac{2}{3}\right)\)

\(=-3\left(x^2-2.\frac{5}{6}.x+\frac{25}{35}-\frac{49}{36}\right)=-3\left[\left(x-\frac{5}{6}\right)^2-\frac{49}{36}\right]=\frac{49}{12}-3\left(x-\frac{5}{6}\right)^2\le\frac{49}{12}\)

Dấu "=" xảy ra khi x=5/6

b)\(D=-8x^2+4xy-y^2+3=3-\left(8x^2-4xy+y^2\right)=3-\left[\left(4x^2-4xy+y^2\right)+4x^2\right]\)

\(=3-\left[\left(2x-y\right)^2+4x^2\right]\le3\)

Dấu "=" xảy ra khi x=y=0

Ta có: A = x² + 2y² + 9z² - 2x + 12y + 6z + 24

A = (x² - 2x + 1) + 2(y² + 6y + 9) + (9z² + 6z + 1) + 4

A = (x - 1)² + 2(y + 3)² + (3z + 1)²  + 4 \(\ge\)4 \(\forall\)x;y;z

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-1=0\\y+3=0\\3z+1=0\end{cases}}\) <=> \(\hept{\begin{cases}x=1\\y=-3\\z=-\frac{1}{3}\end{cases}}\)

Vậy MinA = 4 <=> x=  1 ; y = -3 và z = -1/3

\(x^2+2y^2+9z^2-2x+12y+6z+24\)

\(=\left(x^2-2x+1\right)+\left(9z^2+6z+1\right)+\left(2y^2+12y+22\right)\)

\(=\left(x-1\right)^2+\left(3z+1\right)^2+2\left(y^2+6y+11\right)\)

\(=\left(x-1\right)^2+\left(3z+1\right)^2+2\left(y^2+6y+9+2\right)\)

\(=\left(x-1\right)^2+\left(3z+1\right)^2+2\left(y+3\right)^2+4\ge4\)

Dấu '' = '' xảy ra khi \(\Leftrightarrow\hept{\begin{cases}x-1=0\\3z+1=0\\y+3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\z=-\frac{1}{3}\\y=-3\end{cases}}}\)

Vậy................................