a.
\(A=\left[\dfrac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)^2}+\dfrac{2}{xy}:\left(\dfrac{x+y}{xy}\right)^2\right].\dfrac{1}{x-y}-\left[\dfrac{x^2-y^2+2xy}{\left(x-y\right)\left(x+y\right)^2}+\dfrac{3}{x^2-2x+2}\right]\)
\(=\left[\dfrac{x-y}{x+y}+\dfrac{2xy}{\left(x+y\right)^2}\right].\dfrac{1}{x-y}-\left[\dfrac{x^2-y^2+2xy}{\left(x-y\right)\left(x+y\right)^2}+\dfrac{3}{x^2-2x+2}\right]\)
\(=\left[\dfrac{\left(x-y\right)\left(x+y\right)+2xy}{\left(x+y\right)^2}\right].\dfrac{1}{x-y}-\dfrac{x^2-y^2+2xy}{\left(x-y\right)\left(x+y\right)^2}-\dfrac{3}{x^2-2x+2}\)
\(=\dfrac{x^2-y^2+2xy}{\left(x-y\right)\left(x+y\right)^2}-\dfrac{x^2-y^2+2xy}{\left(x-y\right)\left(x+y\right)^2}-\dfrac{3}{x^2-2x+2}\)
\(=-\dfrac{3}{x^2-2x+2}\)
b.
\(A=-\dfrac{3}{x^2-2x+2}=-\dfrac{3}{\left(x-1\right)^2+1}\ge-\dfrac{3}{0+1}=-3\)
\(A_{min}=-3\) khi \(x=1\) (và y bất kì thỏa mãn ĐKXĐ)
