a.
\(A=\left[\dfrac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)^2}+\dfrac{2}{xy}:\left(\dfrac{x+y}{xy}\right)^2\right].\dfrac{1}{x-y}-\left[\dfrac{x^2-y^2+2xy}{\left(x-y\right)\left(x+y\right)^2}+\dfrac{3}{x^2-2x+2}\right]\)
\(=\left[\dfrac{x-y}{x+2}+\dfrac{2}{xy}.\left(\dfrac{xy}{x+y}\right)^2\right].\dfrac{1}{x-y}-\left[\dfrac{x^2-y^2+2xy}{\left(x-y\right)\left(x+y\right)^2}+\dfrac{3}{x^2-2x+2}\right]\)
\(=\left[\dfrac{x-y}{x+y}+\dfrac{2xy}{\left(x+y\right)^2}\right].\dfrac{1}{x-y}-\left[\dfrac{x^2-y^2+2xy}{\left(x-y\right)\left(x+y\right)^2}+\dfrac{3}{x^2-2x+2}\right]\)
\(=\left[\dfrac{\left(x-y\right)\left(x+y\right)+2xy}{\left(x+y\right)^2}\right].\dfrac{1}{x-y}-\left[\dfrac{x^2-y^2+2xy}{\left(x-y\right)\left(x+y\right)^2}+\dfrac{3}{x^2-2x+2}\right]\)
\(=\dfrac{x^2-y^2+2xy}{\left(x-y\right)\left(x+y\right)^2}-\left[\dfrac{x^2-y^2+2xy}{\left(x-y\right)\left(x+y\right)^2}+\dfrac{3}{x^2-2x+2}\right]\)
\(=\dfrac{-3}{x^2-2x+2}\)
b.
\(A=\dfrac{-3}{\left(x-1\right)^2+1}\)
Do \(\left(x-1\right)^2+1\ge1;\forall x\)
\(\Rightarrow\dfrac{3}{\left(x-1\right)^2+1}\le3;\forall x\)
\(\Rightarrow\dfrac{-3}{\left(x-1\right)^2+1}\ge-3;\forall x\)
Vậy \(A_{min}=-3\) khi \(x=1\) và y bất kì thỏa mãn ĐKXĐ
