Hệ có nghiệm duy nhất: \(\left(\frac{3m+1}{m+1};\frac{m-1}{m+1}\right)\) khi \(m\ne\pm1\)
Lúc này ta có: \(xy=\frac{\left(3m+1\right)\left(m-1\right)}{\left(m+1\right)^2}=\frac{3m^2+m-3m-1}{\left(m+1\right)^2}\)
\(=\frac{3m^2-2m-1}{\left(m+1\right)^2}=\frac{4m^2-\left(m+1\right)^2}{\left(m+1\right)^2}=\)\(\frac{4m^2}{\left(m+1\right)^2}-1\ge-1\)
Dấu "=" xảy ra khi m = 0