\(3x^2+3y^2+6x-12y+15=0\)

\(\Rightarrow3.\left(x^2+y^2+2x-4y+5\right)=0\Rightarrow x^2+y^2+2x-4y+5=0\)

\(\Rightarrow x^2+y^2+2x-4y+1+4=0\)

\(\Rightarrow\left(x^2+2x+1\right)+\left(y^2-4y+4\right)=0\)

\(\Rightarrow\left(x+1\right)^2+\left(y-2\right)^2=0\)

Vì \(\left(x+1\right)^2\ge0;\left(y-2\right)^2\ge0\Rightarrow\left(x+1\right)^2+\left(y-2\right)^2\ge0\)

Mà \(\left(x+1\right)^2+\left(y-2\right)^2=0\)nên để thỏa mãn đẳng thức thì

  \(\left(x+1\right)^2=\left(y-2\right)^2=0\) <=> x=-1 và y=2

\(3x^2+6x+3+3y^2-12y+12=0\)

\(3\left(x^2+2x+1\right)+3\left(y^2-4y+4\right)=0\)

\(3\left(x+1\right)^2+3\left(y-2\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}x+1=0\\y-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-1\\y=2\end{cases}}}\)

b: \(x^2-6x+xy-6y\)

\(=x\left(x-6\right)+y\left(x-6\right)\)

\(=\left(x-6\right)\left(x+y\right)\)

c: \(2x^2+2xy-x-y\)

\(=2x\left(x+y\right)-\left(x+y\right)\)

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

e: \(x^3-3x^2+3x-1=\left(x-1\right)^3\)

Bài 3 :

\(x=3y=2z\)

\(\Rightarrow x=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{2}}\)

\(\Rightarrow\frac{2x}{2}=\frac{3y}{1}=\frac{4z}{2}=\frac{2x-3y+4z}{2-1+2}=\frac{k}{3}\)

\(\Rightarrow x=\frac{k}{3}\)

     \(y=\frac{k}{3}.\frac{1}{3}=\frac{k}{9}\)

     \(z=\frac{k}{3}.\frac{1}{2}=\frac{k}{6}\)

\(x^2+3y^2+2z^2-2x+12y+4z+15=0\)

\(x^2-2x+1+\left(\sqrt{3}y\right)^2+2.6.y+\left(2\sqrt{3}\right)^2+\left(\sqrt{2}z\right)^2+2.2.z+\left(\sqrt{2}\right)^2=0\)

\(\left(x-1\right)^2+\left(\sqrt{3}y+2\sqrt{3}\right)^2+\left(\sqrt{2}z+\sqrt{2}\right)^2=0\)

\(\Rightarrow x=1;y=-2;z=-1\)

<=>(x²-2x+1)+(3y²+12y+12)+(2z²+4z+2)=0

<=>(x-1)²+3(y+2)²+2(z+1)²=0

Vì \(\hept{\begin{cases}\left(x-1\right)^2\ge0\\3\left(y+2\right)^2\ge0\\2\left(z+1\right)^2\ge0\end{cases}\Rightarrow\left(x-1\right)^2+3\left(y+2\right)^2+2\left(z+1\right)^2\ge0}\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-1=0\\y+2=0\\z+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\\z=-1\end{cases}}}\)

\(H=2x^2+9y^2-6xy-6y-12y+2004\)

\(\Rightarrow2H=4x^2+18y^2-12xy-12x-24y+4008\)

             \(=\left(4x^2-12xy+9y^2\right)+9y^2-12x-24y+4008\)

             \(=\left(2x-3y\right)^2-6\left(2x-3y\right)+9+9y^2-42y+49+3950\)

             \(=\left(2x-3y-3\right)^2+\left(3y-7\right)^2+3950\ge3950\)

\(\Rightarrow2H\ge3950\)

\(\Rightarrow H\ge1975\)

Dấu "=" tại \(\hept{\begin{cases}x=5\\y=\frac{7}{3}\end{cases}}\)

\(J=x^2+xy+y^2-3x-3y+1999\)

   \(=\left(x^2+xy+\frac{y^2}{4}\right)+\frac{3y^2}{4}-3x-3y+1999\)

   \(=\left(x+\frac{y}{2}\right)^2-3\left(x+\frac{y}{2}\right)+\frac{9}{4}+3\left(\frac{y^2}{4}-\frac{y}{2}+\frac{1}{4}\right)+1996\)

    \(=\left(x+\frac{y}{2}-\frac{3}{2}\right)^2+3\left(\frac{y}{2}-\frac{1}{2}\right)^2+1996\ge1996\)

Dấu "=" tại \(\hept{\begin{cases}x=1\\y=1\end{cases}}\)

Em có cách này câu J nè:) (tuy nhiên ko gọn như anh, cách này viết thành đa thức giống như đt biến x rồi tìm min thôi)

Ta có: \(J=x^2+2x.\frac{\left(y-3\right)}{2}+y^2-3y+1999\)

\(=x^2+2x.\frac{y-3}{2}+\frac{\left(y-3\right)^2}{4}+y^2-3y+1999-\frac{\left(y-3\right)^2}{4}\)

\(=\left(x+\frac{y-3}{2}\right)^2+\frac{3y^2-6y+3+7984}{4}\)

\(=\left(x+\frac{y-3}{2}\right)^2+\frac{3\left(y-1\right)^2}{4}+1996\ge1996\)