a/ Đặt $A=(x-1)(x-2)(x-3)(x-4)+2003$

$=[(x-1)(x-4)][(x-2)(x-3)]+2003\\=(x^2-5x+4)(x^2-5x+6)+2003$

Đặt $x^2-5x=t$

$→A=(t+4)(t+6)+2003\\=t^2+10t+24+2003\\=t^2+10t+25+2002\\=(t+5)^2+2002$

Ta có: $(t+5)^2\ge 0$

$→A\ge 2002$

$→$ Dấu "=" xảy ra khi $t+5=0$

$↔x^2-5x+5=0\\↔x^2-2.x.\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{5}{4}=0\\↔\left(x-\dfrac{5}{2}\right)^2=\dfrac{5}{4}\\↔\left[\begin{array}{1}x-\dfrac{5}{2}=\dfrac{\sqrt 5}{2}\\x-\dfrac{5}{2}=\dfrac{-\sqrt 5}{2}\end{array}\right.\\↔\left[\begin{array}{1}x=\dfrac{5+\sqrt 5}{2}\\x=\dfrac{5-\sqrt 5}{2}\end{array}\right.$

Vậy $A$ đạt GTNN là $2002$ tại $x=\dfrac{5+\sqrt 5}{2}$ hoặc $x=\dfrac{5-\sqrt 5}{2}$

b/ Đặt $A=3x^2-6xy+5y^2-y+3x+2021$

và $B=(ax+by+c)^2+(dy+e)^2+k$

$B=(ax+by+c)^2+(dy+e)^2+k\\=a^2x^2+b^2y^2+c^2+2abxy+2bcy+2acx+d^2y^2+2dey+e^2+k\\=a^2x^2+(b^2+d^2)y^2+2abxy+2acx+2(bc+de)y+c^2+e^2+k$

Đồng nhất $A$ và $B$

$→\begin{cases}a^2=3\\2ab=-6\\b^2+d^2=5\\2(bc+de)=-1\\2ac=3\\c^2+e^2+k=2021\end{cases}$

$a^2=3\\↔a=\pm\sqrt 3$

Thay $a=\sqrt 3$ vào $2ab=-6$

$2.\sqrt 3.b=-6\\↔\sqrt 3b=-3\\↔b=-\dfrac{3\sqrt 3}{3}$

Thay $a=\sqrt 3$ vào $2ac=3$

$2.\sqrt 3.c=3\\↔c=\dfrac{\sqrt 3}{2}$

Thay $b=-\dfrac{3\sqrt 3}{3}$ vào $b^2+d^2=5$

$\left(-\dfrac{3\sqrt 3}{3}\right)^2+d^2=5\\↔3+d^2=5\\↔d^2=2\\↔d=±\sqrt 2$

Thay $b=-\dfrac{3\sqrt 3}{3},c=\dfrac{\sqrt 3}{2}$ và $d=\sqrt 2$ vào $2(bc+de)=-1$

$2\left(-\dfrac{3\sqrt 3}{3}.\dfrac{\sqrt 3}{2}+\sqrt 2 e\right)=-1\\↔-\dfrac{3}{2}+\sqrt 2 e=-\dfrac{1}{2}\\↔\sqrt 2 e=1\\↔e=\dfrac{\sqrt 2}{2}$

Thay $c=\dfrac{\sqrt 3}{2},e=\dfrac{\sqrt 2}{2}$ vào $c^2+e^2+k=2021$

$\left(\dfrac{\sqrt 3}{2}\right)^2+\left(\dfrac{\sqrt 2}{2}\right)^2+k=2021\\↔\dfrac{5}{4}+k=2021\\↔k=\dfrac{8079}{4}$

Thay $a=\sqrt 3,b=-\dfrac{3\sqrt 3}{3},c=\dfrac{\sqrt 3}{2},d=\sqrt 2,e=\dfrac{\sqrt 2}{2},k=\dfrac{8079}{4}$ vào $B$

$B=\left(\sqrt 3 x-\dfrac{3\sqrt 3}{3}y+\dfrac{\sqrt 3}{2}\right)^2+\left(\sqrt 2 y+\dfrac{\sqrt 2}{2}\right)^2+\dfrac{8079}{4}\\=3\left(x-y+\dfrac{1}{2}\right)+2\left(y+\dfrac{1}{2}\right)^2+\dfrac{8079}{4}\\→A=3\left(x-y+\dfrac{1}{2}\right)^2+2\left(y+\dfrac{1}{2}\right)^2+\dfrac{8079}{4}$

Ta có: $3\left(x-y+\dfrac{1}{2}\right)^2\ge 0,2\left(y+\dfrac{1}{2}\right)^2\ge 0$

$→3\left(x-y+\dfrac{1}{2}\right)^2+2\left(y+\dfrac{1}{2}\right)^2\ge 0\\↔A\ge \dfrac{8079}{4}$

$→$ Dấu "=" xảy ra khi $\begin{cases}x-y+\dfrac{1}{2}=0\\y+\dfrac{1}{2}=0\end{cases}$

$↔\begin{cases}x=y-\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{cases}$

$↔\begin{cases}x=-1\\y=-\dfrac{1}{2}\end{cases}$

Vậy $A$ đạt GTNN là $\dfrac{8079}{4}$ khi $x=-1,y=-\dfrac{1}{2}$