Ta có:x²+3y²-2x+12y+13=0<=>x²+3y²-2x+12y+1+12=0

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

Vì (x-1)²\(\ge0\);3(y+1)²\(\ge0\) nên:(x-1)²+3(y+2)²\(\ge0\)

Dấu "=" xảy ra khi:\(\begin{cases} (x-1)^2=0\\ 3(y+2)^2=0 \end{cases}\)<=>\(\begin{cases} x-1=0\\ y+2=0 \end{cases}\)<=>\(\begin{cases} x=1\\ y=-2 \end{cases}\)

Vậy x=1;y=-2

xin loi mk ghi thieu

x²+3y²-2x+12y+13=0

a. \(x^2+4y^2+z^2=2x+12y-4z-14\)

\(\Leftrightarrow x^2+4y^2+z^2-2x-12y+4z+14=0\)

\(\Leftrightarrow\left(x^2-2x+1\right)+\left(4y^2-12y+9\right)+\left(z^2+4z+4\right)=0\)

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

Ta có: \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\\\left(2y-3\right)^2\ge0\\\left(z+2\right)\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\2y-3=0\\z+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)

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

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\\z+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\\z=-1\end{matrix}\right.\)

\(PT\left(2\right)\Leftrightarrow x=y-1\\ PT\left(1\right)\Leftrightarrow2\left(y-1\right)^2+y\left(1-y\right)+3y^2=7\left(y-1\right)+12y-1\\ \Leftrightarrow2y^2-11y+5=0\\ \Leftrightarrow\left[{}\begin{matrix}y=5\Leftrightarrow x=4\\y=\dfrac{1}{2}\Leftrightarrow x=-\dfrac{1}{2}\end{matrix}\right.\)

Vậy ...

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

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

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

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

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

\(\hept{\begin{cases}x-y+1=0\\2x^2-xy+3y^2-7x-12y+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=y-1\\2x^2-xy+3y^2-7x-12y+1=0\end{cases}}\)

Thế phương trình trên vào phương trình dưới, ta có:

\(2\left(y-1\right)^2-\left(y-1\right)y+3y^2-7\left(y-1\right)-12y+1=0\)\(\Leftrightarrow2y^2-4y+2-y^2+y+3y^2-7y+7-12y+1=0\)

\(\Leftrightarrow4y^2-22y+10=0\Leftrightarrow\orbr{\begin{cases}y=5\\y=\frac{1}{2}\end{cases}}\)

Với y = 5 thì x = 5 - 1 = 4

Với \(y=\frac{1}{2}\Rightarrow x=\frac{1}{2}-1=-\frac{1}{2}\)

Vậy hệ có 2 nghiệm \(\left(4;5\right)\) và \(\left(-\frac{1}{2};\frac{1}{2}\right)\)

Thế y = x + 1 vào phương trình phía dưới.