\(a,A=3x^2-5x+1\)

\(=3\left(x^2-\dfrac{5}{3}x+\dfrac{25}{36}\right)-\dfrac{13}{12}\)

\(=3\left(x-\dfrac{5}{6}\right)^2-\dfrac{13}{12}\)

Với mọi giá trị của x ta có:

\(\left(x-\dfrac{5}{6}\right)^2\ge0\)

\(\Rightarrow3\left(x-\dfrac{5}{6}\right)^2-\dfrac{13}{12}\ge-\dfrac{13}{12}\)

Vậy Min \(A=-\dfrac{13}{12}\)

Để \(A=-\dfrac{13}{12}\) thì \(x-\dfrac{5}{6}=0\Rightarrow x=\dfrac{5}{6}\)

\(b,B=2x^2+5y^2-4x+2y+4xy+2017\)

\(=\left(2x^2-4x+4xy\right)+5y^2+2y+2017\)

\(=2\left(x^2-2x+2xy\right)+5y^2+2y+2017\)

\(=2\left[x^2-2x\left(1-y\right)+\left(1-y\right)^2\right]+5y^2+2y+2017+2\left(1-y\right)^2\)\(=2\left(x-1+y\right)^2+5y^2+2y+2017-2\left(1-y\right)^2\)

\(=2\left(x+y-1\right)^2+5y^2+2y+2017-2+4y-2y^2\)\(=2\left(x+y-1\right)^2+3y^2+6y+2015\)

\(=2\left(x+y-1\right)^2+3\left(y^2+2y+1\right)+2012\)

\(=2\left(x+y-1\right)^2+3\left(y+1\right)^2+2012\)

Với mọi giá trị của x ta có:

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

\(\Rightarrow2\left(x+y-1\right)^2+3\left(y+1\right)^2+2012\ge2012\) Vậy : Min B = 2012

Để B = 2012 thì \(\left\{{}\begin{matrix}x+y-1=0\\y+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)