Giải thích các bước giải:
1,
Ta có:
\(\begin{array}{l}
{a^3} + {b^3} + {c^3} - 3abc = 0\\
\Leftrightarrow {\left( {a + b} \right)^3} - 3ab\left( {a + b} \right) + {c^3} - 3abc = 0\\
\Leftrightarrow \left[ {{{\left( {a + b} \right)}^3} + {c^3}} \right] - 3ab\left( {a + b + c} \right) = 0\\
\Leftrightarrow {\left( {a + b + c} \right)^3} - 3\left( {a + b} \right)c\left( {a + b + c} \right) - 3ab\left( {a + b + c} \right) = 0\\
\Leftrightarrow \left( {a + b + c} \right)\left[ {{{\left( {a + b + c} \right)}^2} - 3\left( {a + b} \right)c - 3ab} \right] = 0\\
\Leftrightarrow \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca - 3ab - 3bc - 3ca} \right) = 0\\
\Leftrightarrow {a^2} + {b^2} + {c^2} - ab - bc - ca = 0\\
\Leftrightarrow \left( {{a^2} - 2ab + {b^2}} \right) + \left( {{b^2} - 2bc + {c^2}} \right) + \left( {{c^2} - 2ca + {a^2}} \right) = 0\\
\Leftrightarrow {\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} = 0\\
\Leftrightarrow a = b = c
\end{array}\)
Vậy tam giác đã cho là tam giác đều.
2,
\(\begin{array}{l}
2{x^2} + 4x + 3{y^2} = 19\\
\Leftrightarrow 2\left( {{x^2} + 2x + 1} \right) + 3{y^2} = 21\\
\Leftrightarrow 2{\left( {x + 1} \right)^2} + 3{y^2} = {2.3^2} + {3.1^2}\\
\Leftrightarrow \left\{ \begin{array}{l}
{\left( {x + 1} \right)^2} = {3^2}\\
{y^2} = {1^2}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x = 2\\
x = - 4
\end{array} \right.\\
\left[ \begin{array}{l}
y = 1\\
y = - 1
\end{array} \right.
\end{array} \right.
\end{array}\)
