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) \(D=x^2-2xy+6y^2-12x+2y+45=\left[x^2-2x\left(y+6\right)+\left(y+6\right)^2\right]+5\left(y^2-2y+1\right)+4=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\ge4\)

\(minD=4\Leftrightarrow\) \(\left\{{}\begin{matrix}x=7\\y=1\end{matrix}\right.\)

Đưa một tỉ tao làm cho 

Ta có: \(C=10x^2+4x-12xy+5y^2+6y+19\)

\(=\left(y^2+6y+9\right)+\left(4y^2-12xy+9x^2\right)+\left(x^2+4x+4\right)+6\)

\(=\left(y+3\right)^2+\left(2y-3x\right)^2+\left(x+2\right)^2+6\ge6\forall x,y\)

Dấu '=' xảy ra khi:

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

Vậy: Giá trị nhỏ nhất của biểu thức \(C=10x^2+4x-12xy+5y^2+6y+19\) là 6 khi x=-2 và y=-3

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,\(5x^2+9y^2-6xy-4x+1=0\)

\(\Rightarrow\left(4x^2-4x+1\right)+\left(x^2-6xy+9y^x\right)=0\)

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

\(\Rightarrow\hept{\begin{cases}2x+1=0\\x-3y=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-\frac{1}{2}\\-\frac{1}{2}-3y=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-\frac{1}{2}\\y=-\frac{1}{6}\end{cases}}\)