Giải thích các bước giải:
ĐK: $a,b,c\ne 0; a,b,c\in Q$
Ta có:
$\begin{array}{l}
\sqrt {\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}} + \dfrac{1}{{{c^2}}}} \\
= \sqrt {\dfrac{1}{{{a^2}}} + \dfrac{{{b^2} + {c^2}}}{{{{\left( {bc} \right)}^2}}}} \\
= \sqrt {\dfrac{1}{{{a^2}}} + \dfrac{{{{\left( {b + c} \right)}^2} - 2bc}}{{{{\left( {bc} \right)}^2}}}} \\
= \sqrt {\dfrac{1}{{{a^2}}} + \dfrac{{{a^2} - 2bc}}{{{{\left( {bc} \right)}^2}}}} \\
= \sqrt {\dfrac{{{{\left( {bc} \right)}^2} + {a^2}\left( {{a^2} - 2bc} \right)}}{{{{\left( {bc} \right)}^2}}}} \\
= \sqrt {\dfrac{{{{\left( {{a^2}} \right)}^2} - 2{a^2}.\left( {bc} \right) + {{\left( {bc} \right)}^2}}}{{{{\left( {bc} \right)}^2}}}} \\
= \sqrt {\dfrac{{{{\left( {{a^2} - bc} \right)}^2}}}{{{{\left( {bc} \right)}^2}}}} \\
= \sqrt {{{\left( {\dfrac{{{a^2} - bc}}{{bc}}} \right)}^2}} \\
= \left| {\dfrac{{{a^2} - bc}}{{bc}}} \right| \in Q
\end{array}$
