Giải thích các bước giải:
Giả sử tam giác có 3 cạnh như đã cho là tam giác ABC với \(a = BC,\,\,b = AC,\,\,c = AB\)
Ta có:
\(\begin{array}{l}
\cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\\
= \frac{{{{\left( {2x + 1} \right)}^2} + {{\left( {{x^2} - 1} \right)}^2} - {{\left( {{x^2} + x + 1} \right)}^2}}}{{2.\left( {2x + 1} \right).\left( {{x^2} - 1} \right)}}\\
= \frac{{{{\left( {2x + 1} \right)}^2} + \left[ {\left( {{x^2} - 1} \right) - \left( {{x^2} + x + 1} \right)} \right]\left[ {\left( {{x^2} - 1} \right) + \left( {{x^2} + x + 1} \right)} \right]}}{{2.\left( {2x + 1} \right).\left( {{x^2} - 1} \right)}}\\
= \frac{{{{\left( {2x + 1} \right)}^2} + \left( { - x - 2} \right).\left( {2{x^2} + x} \right)}}{{2\left( {2x + 1} \right).\left( {{x^2} - 1} \right)}}\\
= \frac{{{{\left( {2x + 1} \right)}^2} - \left( {x - 2} \right).x.\left( {2x + 1} \right)}}{{2.\left( {2x + 1} \right).\left( {{x^2} - 1} \right)}}\\
= \frac{{\left( {2x + 1} \right).\left[ {\left( {2x + 1} \right) - x\left( {x - 2} \right)} \right]}}{{2.\left( {2x + 1} \right)\left( {{x^2} - 1} \right)}}\\
= \frac{{\left( {2x + 1} \right).\left( { - {x^2} + 1} \right)}}{{2.\left( {2x + 1} \right).\left( {{x^2} - 1} \right)}}\\
= - \frac{1}{2}\\
\Rightarrow \widehat A = 120^\circ
\end{array}\)
