A=2x^2+9y^2-6xy-6x-12y+2024 
A = (x^2 -6xy +9y^2) + 4(x -3y) + x^2 - 10x + 2024
A = (x -3y)^2 +4(x -3y) + 4 + x^2 -10x +25 + 1995
A = (x -3y +2)^2 + (x -5)^2 + 1995 \geq 1995
Min A = 1995 
 x - 5 = 0 => x = 5
Và x - 3y + 2 = 0 hay 5 -3y +2 = 0 => -3y = -7 => y = 7/3 

\(K\)\(nha!~!\)

\(\Leftrightarrow\left(x^2-6xy+9y^2\right)+\left(x^2+6x+9\right)+\left(z^2-8z+16\right)=0\)

\(\Leftrightarrow\left(x-3y\right)^2+\left(x+3\right)^2+\left(z-4\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-3y=0\\x+3=0\\z-4=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-1\\z=4\end{matrix}\right.\)

2: \(H=2x^2+4y^2+4xy+4y+9\)

\(=2x^2+4xy+2y^2+2y^2+4y+2+7\)

\(=2\left(x+y\right)^2+2\left(y+1\right)^2+7\ge7\forall x,y\)

Dấu '=' xảy ra khi x+y=0 và y+1=0

=>y=-1 và x=-y=1

3: \(I=x^2-4xy+5y^2+10x-22y+28\)

\(=x^2-4xy+4y^2+10x-20y+y^2-2y+28\)

\(=\left(x-2y\right)^2+10\left(x-2y\right)+25+y^2-2y+1+2\)

\(=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\forall x,y\)

Dấu '=' xảy ra khi y-1=0 và x-2y+5=0

=>y=1 và x=2y-5=2*1-5=-3

4: \(K=x^2+5y^2-4xy+6x-14y+15\)

\(=x^2-4xy+4y^2+6x-12y+y^2-2y+15\)

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

\(=\left(x-2y+3\right)^2+\left(y-1\right)^2+5\ge5\forall x,y\)

Dấu '=' xảy ra khi y-1=0 và x-2y+3=0

=>y=1 và x=2y-3=2*1-3=-1

Ta có \(x^2-6xy+9y^2-3x=0\left(1\right)\)

\(\Leftrightarrow3x=\left(x-3y\right)^2⋮3\Rightarrow3x=\left(x-3y\right)^2⋮9\)

\(\Rightarrow x⋮3\)

Mà \(x\) là số nguyên tố nên \(x=3\)

\(\left(1\right)\Leftrightarrow3x=\left(x-3y\right)^2\)

\(\Leftrightarrow9=\left(9-3y\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}y=2\\y=4\end{matrix}\right.\)

Thử lại được \(x=3;y=2\)

hôm qua mình thi hsg câu này mà ko bt làm 

`9x<sup>2</sup> + 3y<sup>2</sup> + 6xy - 6x + 2y - 35 = 0`

`<=> (9x<sup>2</sup> + 6xy + y<sup>2</sup>) - 2(3x + y) + 1 + 2(y<sup>2</sup> + 2y + 1) - 37 = 0`

`<=> (3x + y - 1)<sup>2</sup> = 37 - 2(y + 1)^2<sup>`</sup>

Vì `(3x+y=1)^2>=0`

`=>2(y+1)^2<=37`

`=>(y+1)^2<=37/2`

Mà `(y+1)^2` là scp

`=>(y+1)^2 in {0,1,4,8,16}`

`=> y + 1 0; 1; -1; 2; -2; 3; -3; 4; -4}`

`=>y in {-1,0,-2,1,-3,2,-4,3,-5}`

Đến đây dễ rồi thay y vào rồi tìm x thôi!

a)\(x^4+3x^3+x^2+3x=x\left(x^3+3x^2+x+3\right)\)

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

b) \(x^2+6xy+9y^2-4z^2=\left(x+3y\right)^2-4z^2=\left(x+3y-2z\right)\left(x+3y+2z\right)\)

c) \(=2x\left(x-1\right)-7\left(x-1\right)=\left(x-1\right)\left(2x-7\right)\)

\(a,=x^3\left(x+3\right)+x\left(x+3\right)=x\left(x^2+1\right)\left(x+3\right)\\ b,=\left(x+3y\right)^2-4z^2=\left(x+3y+2z\right)\left(x+3y-2z\right)\\ c,=2x^2-2x-7x+7=\left(x-1\right)\left(2x-7\right)\)

\(a,4x^2-4x+1\\ =\left(2x\right)^2-2.2x+1^2=\left(2x-1\right)^2\\ c,x^2-6xy-25z^2+9y^2\\ =\left(x^2-2.x.3y+9y^2\right)-\left(5z\right)^2\\ =\left(x-3y\right)^2-\left(5z\right)^2\\ =\left(x-3y-5z\right)\left(x-3y+5z\right)\)

Xem lại đề ý b