k)
\(\begin{array}{l}{c^4} - 2\left( {{a^2} + {b^2}} \right){c^2} + {a^4} + {a^2}{b^2} + {b^4} = 0\\ \Leftrightarrow {c^4} - 2\left( {{a^2} + {b^2}} \right){c^2} + \left( {{a^4} + 2{a^2}{b^2} + {b^4}} \right) - {a^2}{b^2} = 0\\ \Leftrightarrow {c^4} - 2\left( {{a^2} + {b^2}} \right){c^2} + {\left( {{a^2} + {b^2}} \right)^2} - {a^2}{b^2} = 0\\ \Leftrightarrow {\left( {{c^2} - {a^2} - {b^2}} \right)^2} - {a^2}{b^2} = 0\\ \Leftrightarrow \left( {{c^2} - {a^2} - {b^2} - ab} \right)\left( {{c^2} - {a^2} - {b^2} + ab} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}{c^2} - {a^2} - {b^2} - ab = 0\\{c^2} - {a^2} - {b^2} + ab = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}{c^2} = {a^2} + {b^2} + ab\,\,\left( 1 \right)\\{c^2} = {a^2} + {b^2} - ab\,\,\left( 2 \right)\end{array} \right.\end{array}\)
Xét \(\left( 1 \right)\), vì \({c^2} = {a^2} + {b^2} - 2ab\cos C\) nên \({a^2} + {b^2} - 2ab\cos C = {a^2} + {b^2} + ab\)\( \Leftrightarrow \cos C = - \dfrac{1}{2} \Leftrightarrow C = {120^0}\)
Xét \(\left( 2 \right)\), vì \({c^2} = {a^2} + {b^2} - 2ab\cos C\) nên \({a^2} + {b^2} - 2ab\cos C = {a^2} + {b^2} - ab\)\( \Leftrightarrow \cos C = \dfrac{1}{2} \Leftrightarrow C = {60^0}\)
Vậy tam giác ABC có \(C = {60^0}\) hoặc \(C = {120^0}\)
