5) Ta có \(T=\dfrac{x^2+y^2+2}{x-y}=\dfrac{x^2+y^2-2}{x-y}+\dfrac{4}{x-y}=\dfrac{x^2+y^2-2xy}{x-y}+\dfrac{4}{x-y}\)
\(=\dfrac{\left(x-y\right)^2}{x-y}+\dfrac{4}{x-y}=x-y+\dfrac{4}{x-y}\ge2\sqrt{\left(x-y\right).\dfrac{4}{x-y}}=4\)
Dấu "=" xảy ra <=> \(x-y=\dfrac{4}{x-y}\Leftrightarrow x-y=2\)
Kết hợp x.y = 1
=> \(x=\sqrt{2}+1;y=\sqrt{2}-1\)
Vậy Min T = 4 <=> \(x=\sqrt{2}+1;y=\sqrt{2}-1\)