biết \(x^2+3y^2-2x+12y+13\) . khi đó x=... y= .....
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\(x^2-3y^2-2x+12y+13=0\)
\(\Rightarrow\left(x^2-2x+1\right)-3\left(y^2-4y+4\right)+4^2=0\)HÌnh như hơi vô lý bạn ạg
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Bài 3 :
\(x=3y=2z\)
\(\Rightarrow x=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{2}}\)
\(\Rightarrow\frac{2x}{2}=\frac{3y}{1}=\frac{4z}{2}=\frac{2x-3y+4z}{2-1+2}=\frac{k}{3}\)
\(\Rightarrow x=\frac{k}{3}\)
\(y=\frac{k}{3}.\frac{1}{3}=\frac{k}{9}\)
\(z=\frac{k}{3}.\frac{1}{2}=\frac{k}{6}\)
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\(\left(x+3y\right)^2-\left(2x-3y\right)^2-2x^2+12y^2\)
\(=x^2+2\cdot x\cdot3y+\left(3y\right)^2-\left[\left(2x\right)^2-2\cdot2x\cdot3y+\left(3y\right)^2\right]-2x^2+12y^2\)
\(=x^2+6xy+9y^2-4x^2+12xy-9y^2-2x^2+12y^2\)
\(=-5x^2+18xy+12y^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. \(-x^2+4x+y^2-12y+47\)
\(=-\left(x^2-4x-y^2+17y-47\right)\)
\(=-\left[x^2-4x+4-\left(y^2-12y+36\right)-15\right]\)
\(=-\left[\left(x-2\right)^2-\left(y-6\right)^2-15\right]\)
Vì \(\left(x-2\right)^2-\left(y-6\right)^2-15\ge-15\forall x\)
\(\Rightarrow-\left[\left(x-2\right)^2-\left(y-6\right)^2-15\right]\le15\)
Vậy GTLN của bt trên là 15 \(\Leftrightarrow x=2;y=6\)
b. \(-x^2-x-y^2-3y+13\)
\(=\frac{1}{4}\left(-4x^2-4x-4y^2-12y+52\right)\)
\(=\frac{1}{4}\left[-\left(2x+1\right)^2-\left(2y+3\right)^2+42\right]\)
Vì \(\frac{1}{4}\left[-\left(2x+1\right)^2-\left(2y+3\right)^2+42\right]\le42\forall x;y\)
\(\Rightarrow-\left(2x+1\right)^2-\left(2y+3\right)^2+42\le\frac{21}{2}\forall x;y\)
Vậy GTLN của bt trên là 21/2 \(\Leftrightarrow x=-\frac{1}{2};y=-\frac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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. \(x^2+3y^2+2z^2-2x+12y+4z+15=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+3\left(y^2+4y+4\right)+2\left(z^2+2z+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+3\left(y+2\right)^2+2\left(z+1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\\z+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\\z=-1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(x^3-3x^2+3x-1=\left(x-1\right)^3\)
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