Cho hai đường thẳng (d1): \(y = mx + m\) và (d2): \(y =  - \frac{1}{m}x + \frac{1}{m}\) (với m là tham số, \(m \ne 0\)).

Theo đề bài, \(({x_0};{y_0})\) là nghiệm của hệ:

\(\begin{array}{l}
{\rm{    }}\left\{ \begin{array}{l}
{y_0} = m{x_0} + m\\
{y_0} =  - \frac{1}{m}{x_0} + \frac{1}{m}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
m{x_0} + m =  - \frac{1}{m}{x_0} + \frac{1}{m}\\
{y_0} = m{x_0} + m
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{m^2}{x_0} + {m^2} =  - {x_0} + 1\\
{y_0} = m{x_0} + m
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
({m^2} + 1){x_0} = 1 - {m^2}\\
{y_0} = m({x_0} + 1)
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{x_0} = \frac{{1 - {m^2}}}{{1 + {m^2}}}\\
{y_0} = m\left( {\frac{{1 - {m^2}}}{{1 + {m^2}}} + 1} \right)
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{x_0} = \frac{{1 - {m^2}}}{{1 + {m^2}}}\\
{y_0} = \frac{{2m}}{{1 + {m^2}}}
\end{array} \right.
\end{array}\)

Do đó:

\(T = x_0^2 + y_0^2 = {\left( {\frac{{1 - {m^2}}}{{1 + {m^2}}}} \right)^2} + {\left( {\frac{{2m}}{{1 + {m^2}}}} \right)^2} = \frac{{1 - 2{m^2} + {m^4} + 4{m^2}}}{{{{\left( {1 + {m^2}} \right)}^2}}} = \frac{{{{\left( {1 + {m^2}} \right)}^2}}}{{{{\left( {1 + {m^2}} \right)}^2}}} = 1\)