Đáp án: $\begin{array}{l}
a)\,\left( {a;b;c} \right) = \left( {\frac{{14}}{5};\frac{{28}}{5};7} \right)\\
b)\left[ \begin{array}{l}
\left( {a;b;c} \right) = \left( {\frac{{4\sqrt {15} }}{3};2\sqrt {15} ;\frac{{8\sqrt {15} }}{3}} \right)\\
\left( {a;b;c} \right) = \left( {\frac{{ - 4\sqrt {15} }}{3}; - 2\sqrt {15} ;\frac{{ - 8\sqrt {15} }}{3}} \right)
\end{array} \right.
\end{array}$
Giải thích các bước giải:
Áp dụng t/c dãy tỉ số bằng nhau ta có:
$\begin{array}{l}
a)a:b:c = 2:4:5\\
\Rightarrow \frac{a}{2} = \frac{b}{4} = \frac{c}{5} = \frac{{2a}}{4} = \frac{{2a - b + c}}{{4 - 4 + 5}} = \frac{7}{5}\\
\Rightarrow \left\{ \begin{array}{l}
a = \frac{7}{5}.2 = \frac{{14}}{5}\\
b = \frac{7}{5}.4 = \frac{{28}}{5}\\
c = \frac{7}{5}.5 = 7
\end{array} \right.\\
Vay\,\left( {a;b;c} \right) = \left( {\frac{{14}}{5};\frac{{28}}{5};7} \right)\\
b)\frac{a}{2} = \frac{b}{3} = \frac{c}{4}\\
\Rightarrow {\left( {\frac{a}{2}} \right)^2} = {\left( {\frac{b}{3}} \right)^2} = {\left( {\frac{c}{4}} \right)^2}\\
= \frac{{{a^2}}}{4} = \frac{{{b^2}}}{9} = \frac{{{c^2}}}{{16}} = \frac{{2{c^2}}}{{32}} = \frac{{{a^2} - {b^2} + 2{c^2}}}{{4 - 9 + 32}} = \frac{{180}}{{27}} = \frac{{20}}{3}\\
\Rightarrow \left\{ \begin{array}{l}
{a^2} = \frac{{20}}{3}.4 = \frac{{80}}{3}\\
{b^2} = \frac{{20}}{3}.9 = \frac{{180}}{3}\\
{c^2} = \frac{{20}}{3}.16 = \frac{{320}}{3}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\left( {a;b;c} \right) = \left( {\frac{{4\sqrt {15} }}{3};2\sqrt {15} ;\frac{{8\sqrt {15} }}{3}} \right)\\
\left( {a;b;c} \right) = \left( {\frac{{ - 4\sqrt {15} }}{3}; - 2\sqrt {15} ;\frac{{ - 8\sqrt {15} }}{3}} \right)
\end{array} \right.
\end{array}$
