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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
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. \(x^2+3y^2+2z^2-2x+12y+4z+15=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+3\left(y^2+4y+4\right)+2\left(z^2+2z+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+3\left(y+2\right)^2+2\left(z+1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\\z+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\\z=-1\end{matrix}\right.\)
\(x^2+3y^2+2z^2-2x+12y+4z+15=0\)
\(x^2-2x+1+\left(\sqrt{3}y\right)^2+2.6.y+\left(2\sqrt{3}\right)^2+\left(\sqrt{2}z\right)^2+2.2.z+\left(\sqrt{2}\right)^2=0\)
\(\left(x-1\right)^2+\left(\sqrt{3}y+2\sqrt{3}\right)^2+\left(\sqrt{2}z+\sqrt{2}\right)^2=0\)
\(\Rightarrow x=1;y=-2;z=-1\)
<=>(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 <=> \(\hept{\begin{cases}x-1=0\\y+2=0\\z+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\\z=-1\end{cases}}}\)
a, \(A=-x^2+2x+2\)
\(=-\left(x^2-2x-2\right)=-\left(x^2-2x+1-3\right)\)
\(=-\left(x-1\right)^2+3\le3\)
Dấu " = " khi \(-\left(x-1\right)^2=0\Leftrightarrow x=1\)
Vậy \(MAX_A=3\) khi x = 1
b, \(B=-x^2-8x+17\)
\(=-\left(x^2+8x-17\right)\)
\(=-\left(x^2+8x+16-33\right)\)
\(=-\left(x+4\right)^2+33\le33\)
Dấu " = " khi \(-\left(x+4\right)^4=0\Leftrightarrow x=-4\)
Vậy \(MAX_B=33\) khi x = -4
c, \(C=-x^2+7x+15\)
\(=-\left(x^2-\dfrac{7}{2}x.2+\dfrac{49}{4}-\dfrac{109}{4}\right)\)
\(=-\left(x-\dfrac{7}{2}\right)^2+\dfrac{109}{4}\le\dfrac{109}{4}\)
Dấu " = " khi \(-\left(x-\dfrac{7}{2}\right)^2=0\Leftrightarrow x=\dfrac{7}{2}\)
Vậy \(MAX_C=\dfrac{109}{4}\) khi \(x=\dfrac{7}{2}\)
d, \(D=-x^2-5x+11\)
\(=-\left(x^2+\dfrac{5}{2}.x.2+\dfrac{25}{4}-\dfrac{69}{4}\right)\)
\(=-\left(x+\dfrac{5}{2}\right)^2+\dfrac{69}{4}\le\dfrac{69}{4}\)
Dấu " = " khi \(-\left(x+\dfrac{5}{2}\right)^2=0\Leftrightarrow x=\dfrac{-5}{2}\)
Vậy \(MAX_D=\dfrac{69}{4}\) khi \(x=\dfrac{-5}{2}\)
f, sai đề à?
g, \(G=-x^2-x-y^2-3y+13\)
\(=-\left(x^2+x+y^2+3y-13\right)\)
\(=-\left(x^2+\dfrac{1}{2}x.2.+\dfrac{1}{4}+y^2+\dfrac{3}{2}.x.2+\dfrac{9}{4}-15,5\right)\)
\(=-\left(x+\dfrac{1}{2}\right)^2-\left(y+\dfrac{3}{2}\right)^2+15,5\le15,5\)
Dấu " = " khi \(\left\{{}\begin{matrix}-\left(x+\dfrac{1}{2}\right)^2=0\\-\left(y+\dfrac{3}{2}\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-1}{2}\\y=\dfrac{-3}{2}\end{matrix}\right.\)
Vậy \(MAX_G=15,5\) khi \(\left\{{}\begin{matrix}x=\dfrac{-1}{2}\\y=\dfrac{-3}{2}\end{matrix}\right.\)
c: =>(2x+3y-1)^2+(2x-3y)=0
=>2x-3y=0 và 2x+3y=1
=>x=1/4; y=1/6
d: =>2y-3=0 và 2x+3y-1=0
=>y=3/2 và 2x=1-3y=1-9/2=-7/2
=>x=-7/4 và y=3/2
\(P=\left(4x^2-4xy+y^2\right)+\left(x^2+2x+1\right)+3y^2-12y+2026\)
\(P=\left(2y-x-3\right)^2+\left(2x-1\right)^2+2026\)
Pmin =2026 khi x=1/2; y=7/4
\(H=2x^2+9y^2-6xy-6y-12y+2004\)
\(\Rightarrow2H=4x^2+18y^2-12xy-12x-24y+4008\)
\(=\left(4x^2-12xy+9y^2\right)+9y^2-12x-24y+4008\)
\(=\left(2x-3y\right)^2-6\left(2x-3y\right)+9+9y^2-42y+49+3950\)
\(=\left(2x-3y-3\right)^2+\left(3y-7\right)^2+3950\ge3950\)
\(\Rightarrow2H\ge3950\)
\(\Rightarrow H\ge1975\)
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 \(\hept{\begin{cases}x=1\\y=1\end{cases}}\)
\(x^2+3y^2+2z^2-2x+12y+4z+15=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+3\left(y^2+4y+4\right)+2\left(z^2+2z+1\right)=0\)
ta có : \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\forall x\\\left(y+2\right)^2\ge0\forall y\\\left(z+1\right)^2\ge0\forall z\end{matrix}\right.\) \(\Rightarrow\) \(\left(x^2-2x+1\right)+3\left(y^2+4y+4\right)+2\left(z^2+2z+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\\\left(z+1\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\\z+1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\\z=-1\end{matrix}\right.\)
vậy \(x=1;y=-2;z=-1\)
\(x^2+3y^2+2z^2-2z+12y+4z+15=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(3y^2+12y+12\right)+\left(2z^2-4z+2\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+3\left(y+4\right)^2+2\left(z-2\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-4\\z=2\end{matrix}\right.\)
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