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a) \(C=4x^2+3y^2+4xy-4x-10y+7=\left[4x^2+4x\left(y-1\right)+\left(y-1\right)^2\right]+2\left(y^2-4y+4\right)-2=\left(2x+y-1\right)^2+2\left(y-2\right)^2-2\ge-2\)
\(minC=-2\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=2\end{matrix}\right.\)
d) \(D=x^2-2xy+6y^2-12x+2y+45=\left[x^2-2x\left(y+6\right)+\left(y+6\right)^2\right]+5\left(y^2-2y+1\right)+4=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\ge4\)
\(minD=4\Leftrightarrow\) \(\left\{{}\begin{matrix}x=7\\y=1\end{matrix}\right.\)
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a, \(x^2+y^2-2x+10y+26=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2+10y+25\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y+5\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x-1=0\\y+5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-5\end{cases}}\)
b,\(4x^2+2y^2+2xy-2y+1=0\)
\(\Leftrightarrow\left(4x^2+4xy+y^2\right)+\left(y^2-2y+1\right)=0\)
\(\Leftrightarrow\left(2x+y\right)^2+\left(y-1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}2x+y=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x+1=0\\y=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{2}\\y=1\end{cases}}\)
c,\(5x^2+9y^2-12xy+4x+4=0\)
\(\Rightarrow\left(x^2+4x+4\right)+\left(4x^2-12xy+9y^2\right)=0\)
\(\Rightarrow\left(x+2\right)^2+\left(2x-3y\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}x+2=0\\2x-3y=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-2\\2.\left(-2\right)-3y=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-2\\y=-\frac{4}{3}\end{cases}}\)
d,\(5x^2+9y^2-6xy-4x+1=0\)
\(\Rightarrow\left(4x^2-4x+1\right)+\left(x^2-6xy+9y^x\right)=0\)
\(\Rightarrow\left(2x+1\right)^2+\left(x-3y\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}2x+1=0\\x-3y=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-\frac{1}{2}\\-\frac{1}{2}-3y=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-\frac{1}{2}\\y=-\frac{1}{6}\end{cases}}\)
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Lời giải:
Đặt biểu thức trên là $A$
$A=4x^2+4x-12xy-2y+10y^2+8$
$=(4x^2-12xy+9y^2)+4x-2y+y^2+8$
$=(2x-3y)^2+2(2x-3y)+4y+y^2+8$
$=(2x-3y)^2+2(2x-3y)+1+(y^2+4y+4)+3$
$=(2x-3y+1)^2+(y+2)^2+3\geq 0+0+3=3$
Vậy $A_{\min}=3$. Giá trị này đạt tại $2x-3y+1=y+2=0$
$\Leftrightarrow y=-2; x=\frac{-7}{2}$