Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a + b + c = 1 \Rightarrow \left\{ \begin{array}{l}
a + b = 1 - c\\
b + c = 1 - a\\
c + a = 1 - b
\end{array} \right.\\
{a^2} + {b^2} + abc - 1 = {\left( {a + b} \right)^2} - 2ac + abc - 1 = {\left( {1 - c} \right)^2} - 2ab + abc - 1\\
= 1 - 2c + {c^2} - 2ab + abc - 1 = {c^2} - 2c + abc - 2ab = c\left( {c - 2} \right) + ab\left( {c - 2} \right)\\
= \left( {c + ab} \right)\left( {c - 2} \right)\\
{b^2} + {c^2} + abc - 1 = {\left( {b + c} \right)^2} - 2bc + abc - 1 = {\left( {1 - a} \right)^2} - 2bc + abc - 1\\
= 1 - 2a + {a^2} - 2bc + abc - 1 = {a^2} - 2a + abc - 2bc = a\left( {a - 2} \right) + bc\left( {a - 2} \right)\\
= \left( {a + bc} \right)\left( {a + 2} \right)\\
{c^2} + {a^2} + abc - 1 = {\left( {c + a} \right)^2} - 2ca + abc - 1 = {\left( {1 - b} \right)^2} - 2ca + abc - 1\\
= 1 - 2b + {b^2} - 2ca + abc - 1 = {b^2} - 2b + abc - 2ca = b\left( {b - 2} \right) + ca\left( {b - 2} \right)\\
= \left( {b + ca} \right)\left( {b - 2} \right)\\
\Rightarrow \frac{{c + ab}}{{{a^2} + {b^2} + abc - 1}} + \frac{{a + bc}}{{{b^2} + {c^2} + abc - 1}} + \frac{{b + ac}}{{{a^2} + {c^2} + abc - 1}}\\
= \frac{{c + ab}}{{\left( {c + ab} \right)\left( {c - 2} \right)}} + \frac{{a + bc}}{{\left( {a + bc} \right)\left( {a - 2} \right)}} + \frac{{b + ac}}{{\left( {b + ac} \right)\left( {c - 2} \right)}}\\
= \frac{1}{{c - 2}} + \frac{1}{{a - 2}} + \frac{1}{{b - 2}}\\
= \frac{{\left( {a - 2} \right)\left( {b - 2} \right) + \left( {b - 2} \right)\left( {c - 2} \right) + \left( {c - 2} \right)\left( {a - 2} \right)}}{{\left( {a - 2} \right)\left( {b - 2} \right)\left( {c - 2} \right)}}\\
= \frac{{ab - 2a - 2b + 4 + bc - 2b - 2c + 4 + ca - 2c - 2a + 4}}{{\left( {a - 2} \right)\left( {b - 2} \right)\left( {c - 2} \right)}}\\
= \frac{{\left( {ab + bc + ca} \right) - 4.\left( {a + b + c} \right) + 4.3}}{{\left( {a - 2} \right)\left( {b - 2} \right)\left( {c - 2} \right)}}\\
= \frac{{ab + bc + ca - 4.1 + 12}}{{\left( {a - 2} \right)\left( {b - 2} \right)\left( {c - 2} \right)}}\\
= \frac{{ab + bc + ca + 8}}{{\left( {a - 2} \right)\left( {b - 2} \right)\left( {c - 2} \right)}}
\end{array}\)
