Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{\left( {\frac{{{q^4} - 1}}{{q - 1}}} \right)^2}:\frac{{{q^8} - 1}}{{{q^2} - 1}} = \frac{{225}}{{85}}\\
\Leftrightarrow {\left( {\frac{{\left( {{q^2} - 1} \right)\left( {{q^2} + 1} \right)}}{{q - 1}}} \right)^2}:\frac{{\left( {{q^4} - 1} \right)\left( {{q^4} + 1} \right)}}{{{q^2} - 1}} = \frac{{45}}{{17}}\\
\Leftrightarrow {\left( {\frac{{\left( {q - 1} \right)\left( {q + 1} \right)\left( {{q^2} + 1} \right)}}{{q - 1}}} \right)^2}:\frac{{\left( {{q^2} - 1} \right)\left( {{q^2} + 1} \right)\left( {{q^4} + 1} \right)}}{{{q^2} - 1}} = \frac{{45}}{{17}}\\
\Leftrightarrow \frac{{{{\left( {q + 1} \right)}^2}{{\left( {{q^2} + 1} \right)}^2}}}{{\left( {{q^2} + 1} \right)\left( {{q^4} + 1} \right)}} = \frac{{45}}{{17}}\\
\Leftrightarrow \frac{{\left( {{q^2} + 2q + 1} \right)\left( {{q^2} + 1} \right)}}{{{q^4} + 1}} = \frac{{45}}{{17}}\\
\Leftrightarrow \frac{{{q^4} + {q^2} + 2{q^3} + 2q + {q^2} + 1}}{{{q^4} + 1}} = \frac{{45}}{{17}}\\
\Leftrightarrow \frac{{{q^4} + 2{q^3} + 2{q^2} + 2q + 1}}{{{q^4} + 1}} = \frac{{45}}{{17}}\\
\Leftrightarrow 17{q^4} + 34{q^3} + 34{q^2} + 34q + 17 = 45{q^4} + 45\\
\Leftrightarrow 28{q^4} - 34{q^3} - 34{q^2} - 34q + 28 = 0\\
\Leftrightarrow 14{q^4} - 17{q^3} - 17{q^2} - 17q + 14 = 0\\
\Leftrightarrow \left( {14{q^4} - 7{q^3}} \right) - \left( {10{q^3} - 5{q^2}} \right) - \left( {22{q^2} - 11q} \right) - \left( {28q - 14} \right) = 0\\
\Leftrightarrow \left( {2q - 1} \right)\left( {7{q^3} - 10{q^2} - 11q - 14} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
q = \frac{1}{2}\\
q = 2,42..
\end{array} \right.
\end{array}\)
