\(2T=\frac{a^2-2ac+c^2+c^2-2bc+b^2+a^2-2ab+b^2}{\left(a-c\right)\left(a+c\right)-2b\left(a-c\right)}\)
\(2T=\frac{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}{\left(a-c\right)\left(a-b+c-b\right)}\)
Theo đề bài ta có:\(\hept{\begin{cases}a-b=4\\b-c=2\end{cases}\Rightarrow}a-c=6\)
\(\Rightarrow2T=\frac{4^2+2^2+6^2}{6\cdot\left(4-2\right)}=\frac{14}{3}\)