Gọi  \(B\left( {a;2 + \frac{2}{{a - 1}}} \right),C\left( {c;2 + \frac{2}{{c - 1}}} \right)\left( {a < 1 < c} \right)\)

Gọi H, K lần lượt là hình chiếu của B, C trên trục \($Ox \Rightarrow H\left( {a,0} \right),K(c;0)\)

Tam giác ABC vuông cân \(Ox \Rightarrow H\left( {a,0} \right),K(c;0)\)

Ta có: \(\angle BCA = \angle CAK + \angle ACK = \angle BAH + \angle ABH\)

Mà: \(\angle BAH + \angle CAK = {90^0}\)

\( \Rightarrow \angle BAH = \angle ACK\)

Xét \(\Delta ABH\) và \(\Delta CAK\) ta có:

\(\begin{array}{l}
\angle BAH = \angle ACK\,(CMT)\\
AC = AB\,\,\,(gt)\\
 =  > \Delta ABH = \angle CAK\,\,(ch - gn)
\end{array}\)

\( =  > AH = CK,HB = AK\) (các cạnh tương ứng bằng nhau)

Ta có:  \(AH\left| {a - 2} \right| = 2 - a;AK = \left| {c - 2} \right|;\left( {a < 1} \right)\)

\(\begin{array}{l}
BH = \left| {2 + \frac{2}{{a + 1}}} \right|;CK = \left| {2 + \frac{2}{{c - 1}}} \right| = 2 + \frac{2}{{c - 1}}(c > 1)\\
 \Rightarrow \left\{ {\begin{array}{*{20}{c}}
{AH = CK}\\
{HB = AK}
\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}
{2 - a = 2 + \frac{2}{{c - 1}}}\\
{\left| {2 + \frac{2}{{a - 1}}} \right| = \left| {c - 2} \right|}
\end{array}} \right.\\
 \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}
{a = \frac{2}{{1 - c}}}\\
{\left[ {\begin{array}{*{20}{c}}
{2 + \frac{2}{{a - 1}} = c - 2}\\
{2 + \frac{2}{{a - 1}} = 2 - c}
\end{array}} \right.}
\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}
{a = \frac{2}{{1 - c}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\
{\left[ {\begin{array}{*{20}{c}}
{4 + \frac{2}{{\frac{2}{{1 - c}} - 1}} = c - 2}\\
{c = \frac{2}{{1 - a}} = \frac{2}{{1 - \frac{2}{{1 - c}}}}}
\end{array}} \right.}
\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}
{b =  - 1\,\,\,(tm)}\\
{c = 3\,\,(tm)}
\end{array}} \right.\\
 \Rightarrow \left\{ {\begin{array}{*{20}{c}}
{B\left( { - 1;1} \right)}\\
{C\left( {3;3} \right)}
\end{array}} \right. =  > T = \left( { - 1} \right).1 + 3.3 = 8
\end{array}\)