Giải thích các bước giải:
\(\begin{array}{l}
\sqrt {{x_1}^2 + 2018} - {x_1} = \sqrt {{x_2}^2 + 2018} + {x_2}\\
\to \sqrt {{x_1}^2 + 2018} - \sqrt {{x_2}^2 + 2018} = {x_1} + {x_2}\\
\to \dfrac{{\left( {\sqrt {{x_1}^2 + 2018} - \sqrt {{x_2}^2 + 2018} } \right)\left( {\sqrt {{x_1}^2 + 2018} + \sqrt {{x_2}^2 + 2018} } \right)}}{{\sqrt {{x_1}^2 + 2018} + \sqrt {{x_2}^2 + 2018} }} = {x_1} + {x_2}\\
\to \dfrac{{{{\left( {\sqrt {{x_1}^2 + 2018} } \right)}^2} - {{\left( {\sqrt {{x_2}^2 + 2018} } \right)}^2}}}{{\sqrt {{x_1}^2 + 2018} + \sqrt {{x_2}^2 + 2018} }} = {x_1} + {x_2}\\
\to \dfrac{{{x_1}^2 + 2018 - {x_2}^2 - 2018}}{{\sqrt {{x_1}^2 + 2018} + \sqrt {{x_2}^2 + 2018} }} = {x_1} + {x_2}\\
\to \dfrac{{{x_1}^2 - {x_2}^2}}{{\sqrt {{x_1}^2 + 2018} + \sqrt {{x_2}^2 + 2018} }} - \left( {{x_1} + {x_2}} \right) = 0\\
\to \dfrac{{\left( {{x_1} + {x_2}} \right)\left( {{x_1} - {x_2}} \right)}}{{\sqrt {{x_1}^2 + 2018} + \sqrt {{x_2}^2 + 2018} }} - \left( {{x_1} + {x_2}} \right) = 0\\
\to \left( {{x_1} + {x_2}} \right)\left( {\dfrac{{{x_1} - {x_2}}}{{\sqrt {{x_1}^2 + 2018} + \sqrt {{x_2}^2 + 2018} }} - 1} \right) = 0\\
\to \left[ \begin{array}{l}
{x_1} + {x_2} = 0\\
\dfrac{{{x_1} - {x_2}}}{{\sqrt {{x_1}^2 + 2018} + \sqrt {{x_2}^2 + 2018} }} - 1 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
{x_1} + {x_2} = 0\\
{x_1} - {x_2} = \sqrt {{x_1}^2 + 2018} + \sqrt {{x_2}^2 + 2018}
\end{array} \right.
\end{array}\)
( bên dưới b làm tương tự như trong ảnh áp dụng Vi-et )
