Đáp án:
$\begin{array}{l}
a)x + 2019\sqrt {x - 2} = 2\sqrt {x - 1} \left( {dk:x \ge 2} \right)\\
\Rightarrow x - 2\sqrt {x - 1} + 2019\sqrt {x - 2} = 0\\
\Rightarrow \left( {x - 1} \right) - 2\sqrt {x - 1} + 1 + 2019\sqrt {x - 2} = 0\\
\Rightarrow {\left( {\sqrt {x - 1} - 1} \right)^2} + 2019\sqrt {x - 2} = 0\\
Do:{\left( {\sqrt {x - 1} - 1} \right)^2};2019\sqrt {x - 2} \ge 0\forall x \ge 2\\
\Rightarrow \left\{ \begin{array}{l}
\sqrt {x - 1} = 1\\
\sqrt {x - 2} = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x - 1 = 1\\
x - 2 = 0
\end{array} \right.\\
\Rightarrow x = 2\left( {tmdk} \right)\\
Vậy\,x = 2\\
b)x + y + xy = \dfrac{5}{4}\\
\Rightarrow xy = \dfrac{5}{4} - \left( {x + y} \right)\\
A = {x^2} + {y^2}\\
= {\left( {x + y} \right)^2} - 2xy\\
= {\left( {x + y} \right)^2} - 2.\left( {\dfrac{5}{4} - \left( {x + y} \right)} \right)\\
= {\left( {x + y} \right)^2} + 2\left( {x + y} \right) - \dfrac{5}{2}\\
= {\left( {x + y} \right)^2} + 2\left( {x + y} \right) + 1 - \dfrac{7}{2}\\
= {\left( {x + y + 1} \right)^2} - \dfrac{7}{2} \ge - \dfrac{7}{2}\forall x;y\\
\Rightarrow GTNN:A = - \dfrac{7}{2}
\end{array}$
