Đáp án:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
$\begin{array}{l}
a)3 - 2b + c = 24\\
\Rightarrow - 2b + c = 21\\
Do:5a = 8b = 8c\\
\Rightarrow \dfrac{{5a}}{{40}} = \dfrac{{8b}}{{40}} = \dfrac{{8c}}{{40}}\\
= \dfrac{a}{8} = \dfrac{b}{5} = \dfrac{c}{5} = \dfrac{{2b}}{{10}}\\
= \dfrac{{ - 2b + c}}{{ - 10 + 5}} = \dfrac{{21}}{{ - 5}} = \dfrac{{ - 21}}{5}\\
\Rightarrow \left\{ \begin{array}{l}
a = - \dfrac{{21}}{5}.8 = \dfrac{{ - 168}}{5}\\
b = c = \dfrac{{ - 21}}{5}.5 = - 21
\end{array} \right.\\
\text{Vậy}\,a = \dfrac{{ - 168}}{5};b = c = - 21\\
b)3a = 7b\\
\Rightarrow \dfrac{a}{7} = \dfrac{b}{3}\\
\Rightarrow {\left( {\dfrac{a}{7}} \right)^2} = {\left( {\dfrac{b}{3}} \right)^2} = \dfrac{{{a^2}}}{{49}} = \dfrac{{{b^2}}}{9}\\
= \dfrac{{{a^2} - {b^2}}}{{49 - 9}} = \dfrac{{160}}{{40}} = 4\\
\Rightarrow \left\{ \begin{array}{l}
{a^2} = 4.49\\
{b^2} = 4.9
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
a = 14;b = 6\\
a = - 14;b = - 6
\end{array} \right.\\
\text{Vậy}\,\left( {a;b} \right) = \left( {14;6} \right)/\left( { - 14; - 6} \right)\\
c)\dfrac{a}{b} = \dfrac{c}{a} = k \Rightarrow \left\{ \begin{array}{l}
a = b.k\\
c = a.k = b.k.k = b.{k^2}
\end{array} \right.\\
\Rightarrow {a^2} = b.c\\
+ )\dfrac{{{b^2} - {c^2}}}{{{a^2} + {c^2}}} = \dfrac{{{b^2} - {{\left( {b.{k^2}} \right)}^2}}}{{{{\left( {b.k} \right)}^2} + {{\left( {b.{k^2}} \right)}^2}}}\\
= \dfrac{{{b^2} - {b^2}.{k^4}}}{{{b^2}{k^2} + {b^2}{k^4}}} = \dfrac{{1 - {k^4}}}{{{k^2} + {k^4}}}\\
= \dfrac{{\left( {1 - {k^2}} \right)\left( {1 + {k^2}} \right)}}{{{k^2}\left( {1 + {k^2}} \right)}}\\
= \dfrac{{1 - {k^2}}}{{{k^2}}}\\
+ \dfrac{{b - c}}{c} = \dfrac{{b - b.{k^2}}}{{b.{k^2}}} = \dfrac{{1 - {k^2}}}{{{k^2}}}\\
\text{Vậy}\,\dfrac{{{b^2} - {c^2}}}{{{a^2} + {c^2}}} = \dfrac{{b - c}}{c}
\end{array}$
