Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\dfrac{1}{a} + \dfrac{1}{b} = \dfrac{1}{{2018}}\\
\Leftrightarrow \dfrac{{a + b}}{{ab}} = \dfrac{1}{{2018}}\\
\Leftrightarrow 2018.\left( {a + b} \right) = ab\\
\Rightarrow ab - 2018\left( {a + b} \right) = 0\\
V{P^2} = {\left( {\sqrt {a - 2018} + \sqrt {b - 2018} } \right)^2}\\
= \left( {a - 2018} \right) + 2.\sqrt {a - 2018} .\sqrt {b - 2018} + \left( {b - 2018} \right)\\
= \left( {a + b} \right) - 2.2018 + 2.\sqrt {\left( {a - 2018} \right)\left( {b - 2018} \right)} \\
= \left( {a + b} \right) - 2.2018 + 2.\sqrt {ab - 2018.\left( {a + b} \right) + {{2018}^2}} \\
= \left( {a + b} \right) - 2.2018 + 2.\sqrt {0 + {{2018}^2}} \\
= \left( {a + b} \right) - 2.2018 + 2.2018\\
= a + b = V{T^2}\\
\Rightarrow \sqrt {a - 2018} + \sqrt {b - 2018} = a + b
\end{array}\)
