\(E=5x^2+y^2+10+4xy-14x-6y\)

\(E=\left(2x+y-3\right)^2+\left(x-1\right)^2+6\)

Vì \(\left(2x+y-3\right)^2+\left(x-1\right)^2\ge0\)

Dấu '=" xảy ra.......................

\(E=5x^2+y^2+10+4xy-14x-6y\)

\(=\left(4x^2+y^2+4xy\right)-12x-6y+9+x^2-2y+1\)

\(=\left(2x+y\right)^2-6\left(2x+y\right)+9+\left(x-1\right)^2\)

\(=\left(2x+y-3\right)^2+\left(x-1\right)^2\ge0\)

\(\Rightarrow E_{Min}=0\)

\("="\Leftrightarrow x=y=1\)

Ta có E= \(\left(4x^2+y^2+9-6y-12x+4xy\right)+\left(x^2-2x+1\right)\)

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

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

=>E>=0 =>GTNN của E=0 khi: \(x-1=0\) =>\(x=1\)

\(2x+y-3=0\) =>\(2x+y=3\)

=> \(2+y=3\) => \(y=1\)

a)

\(A=2x^2-3x+1=2\left(x^2-\frac{3}{2}x+\frac{9}{16}\right)-2.\frac{9}{16}+1=2\left(x-\frac{3}{4}\right)^2-\frac{1}{8}\ge-\frac{1}{8}\)

Vậy \(MinA=-\frac{1}{8}\Leftrightarrow\left(x-\frac{3}{4}\right)^2=0\Leftrightarrow x=\frac{3}{4}\)

b)

\(B=5x^2+y^2+10+4xy-15x-6y\)

\(=\left[\left(2x\right)^2+y^2-3^2+2.2x.y-2.y.3-2.2x.3\right]+\left(x^2-3x+\frac{9}{4}\right)+\frac{27}{4}\)

\(=\left(2x+y-3\right)^2+\left(x-\frac{3}{2}\right)^2+\frac{27}{4}\ge\frac{27}{4}\)

Vậy \(MinB=\frac{27}{4}\Leftrightarrow\hept{\begin{cases}\left(2x+y-3\right)^2=0\\\left(x-\frac{3}{2}\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}2x+y-3=0\\x-\frac{3}{2}=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{3}{2}\\y=0\end{cases}}}\)

A là -0,125