giải phương trình sau (q^4-1/q-1)^2/(q^8-1/q^2-1)=225/85 13/07/2021 Bởi Anna giải phương trình sau (q^4-1/q-1)^2/(q^8-1/q^2-1)=225/85
Đáp án: Giải thích các bước giải: Ta có: (q4−1q−1)2:q8−1q2−1=22585⇔((q2−1)(q2+1)q−1)2:(q4−1)(q4+1)q2−1=4517⇔((q−1)(q+1)(q2+1)q−1)2:(q2−1)(q2+1)(q4+1)q2−1=4517⇔(q+1)2(q2+1)2(q2+1)(q4+1)=4517⇔(q2+2q+1)(q2+1)q4+1=4517⇔q4+q2+2q3+2q+q2+1q4+1=4517⇔q4+2q3+2q2+2q+1q4+1=4517⇔17q4+34q3+34q2+34q+17=45q4+45⇔28q4−34q3−34q2−34q+28=0⇔14q4−17q3−17q2−17q+14=0⇔(14q4−7q3)−(10q3−5q2)−(22q2−11q)−(28q−14)=0⇔(2q−1)(7q3−10q2−11q−14)=0⇔[q=12q=2,42.. Bình luận
Giải thích các bước giải: Ta có: \(\begin{array}{l}{\left( {\frac{{{q^4} – 1}}{{q – 1}}} \right)^2}:\frac{{{q^8} – 1}}{{{q^2} – 1}} = \frac{{225}}{{85}}\\ \Leftrightarrow {\left( {\frac{{\left( {{q^2} – 1} \right)\left( {{q^2} + 1} \right)}}{{q – 1}}} \right)^2}:\frac{{\left( {{q^4} – 1} \right)\left( {{q^4} + 1} \right)}}{{{q^2} – 1}} = \frac{{45}}{{17}}\\ \Leftrightarrow {\left( {\frac{{\left( {q – 1} \right)\left( {q + 1} \right)\left( {{q^2} + 1} \right)}}{{q – 1}}} \right)^2}:\frac{{\left( {{q^2} – 1} \right)\left( {{q^2} + 1} \right)\left( {{q^4} + 1} \right)}}{{{q^2} – 1}} = \frac{{45}}{{17}}\\ \Leftrightarrow \frac{{{{\left( {q + 1} \right)}^2}{{\left( {{q^2} + 1} \right)}^2}}}{{\left( {{q^2} + 1} \right)\left( {{q^4} + 1} \right)}} = \frac{{45}}{{17}}\\ \Leftrightarrow \frac{{\left( {{q^2} + 2q + 1} \right)\left( {{q^2} + 1} \right)}}{{{q^4} + 1}} = \frac{{45}}{{17}}\\ \Leftrightarrow \frac{{{q^4} + {q^2} + 2{q^3} + 2q + {q^2} + 1}}{{{q^4} + 1}} = \frac{{45}}{{17}}\\ \Leftrightarrow \frac{{{q^4} + 2{q^3} + 2{q^2} + 2q + 1}}{{{q^4} + 1}} = \frac{{45}}{{17}}\\ \Leftrightarrow 17{q^4} + 34{q^3} + 34{q^2} + 34q + 17 = 45{q^4} + 45\\ \Leftrightarrow 28{q^4} – 34{q^3} – 34{q^2} – 34q + 28 = 0\\ \Leftrightarrow 14{q^4} – 17{q^3} – 17{q^2} – 17q + 14 = 0\\ \Leftrightarrow \left( {14{q^4} – 7{q^3}} \right) – \left( {10{q^3} – 5{q^2}} \right) – \left( {22{q^2} – 11q} \right) – \left( {28q – 14} \right) = 0\\ \Leftrightarrow \left( {2q – 1} \right)\left( {7{q^3} – 10{q^2} – 11q – 14} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}q = \frac{1}{2}\\q = 2,42..\end{array} \right.\end{array}\) Bình luận
Đáp án:
Giải thích các bước giải:
Ta có:
(q4−1q−1)2:q8−1q2−1=22585⇔((q2−1)(q2+1)q−1)2:(q4−1)(q4+1)q2−1=4517⇔((q−1)(q+1)(q2+1)q−1)2:(q2−1)(q2+1)(q4+1)q2−1=4517⇔(q+1)2(q2+1)2(q2+1)(q4+1)=4517⇔(q2+2q+1)(q2+1)q4+1=4517⇔q4+q2+2q3+2q+q2+1q4+1=4517⇔q4+2q3+2q2+2q+1q4+1=4517⇔17q4+34q3+34q2+34q+17=45q4+45⇔28q4−34q3−34q2−34q+28=0⇔14q4−17q3−17q2−17q+14=0⇔(14q4−7q3)−(10q3−5q2)−(22q2−11q)−(28q−14)=0⇔(2q−1)(7q3−10q2−11q−14)=0⇔[q=12q=2,42..
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{\left( {\frac{{{q^4} – 1}}{{q – 1}}} \right)^2}:\frac{{{q^8} – 1}}{{{q^2} – 1}} = \frac{{225}}{{85}}\\
\Leftrightarrow {\left( {\frac{{\left( {{q^2} – 1} \right)\left( {{q^2} + 1} \right)}}{{q – 1}}} \right)^2}:\frac{{\left( {{q^4} – 1} \right)\left( {{q^4} + 1} \right)}}{{{q^2} – 1}} = \frac{{45}}{{17}}\\
\Leftrightarrow {\left( {\frac{{\left( {q – 1} \right)\left( {q + 1} \right)\left( {{q^2} + 1} \right)}}{{q – 1}}} \right)^2}:\frac{{\left( {{q^2} – 1} \right)\left( {{q^2} + 1} \right)\left( {{q^4} + 1} \right)}}{{{q^2} – 1}} = \frac{{45}}{{17}}\\
\Leftrightarrow \frac{{{{\left( {q + 1} \right)}^2}{{\left( {{q^2} + 1} \right)}^2}}}{{\left( {{q^2} + 1} \right)\left( {{q^4} + 1} \right)}} = \frac{{45}}{{17}}\\
\Leftrightarrow \frac{{\left( {{q^2} + 2q + 1} \right)\left( {{q^2} + 1} \right)}}{{{q^4} + 1}} = \frac{{45}}{{17}}\\
\Leftrightarrow \frac{{{q^4} + {q^2} + 2{q^3} + 2q + {q^2} + 1}}{{{q^4} + 1}} = \frac{{45}}{{17}}\\
\Leftrightarrow \frac{{{q^4} + 2{q^3} + 2{q^2} + 2q + 1}}{{{q^4} + 1}} = \frac{{45}}{{17}}\\
\Leftrightarrow 17{q^4} + 34{q^3} + 34{q^2} + 34q + 17 = 45{q^4} + 45\\
\Leftrightarrow 28{q^4} – 34{q^3} – 34{q^2} – 34q + 28 = 0\\
\Leftrightarrow 14{q^4} – 17{q^3} – 17{q^2} – 17q + 14 = 0\\
\Leftrightarrow \left( {14{q^4} – 7{q^3}} \right) – \left( {10{q^3} – 5{q^2}} \right) – \left( {22{q^2} – 11q} \right) – \left( {28q – 14} \right) = 0\\
\Leftrightarrow \left( {2q – 1} \right)\left( {7{q^3} – 10{q^2} – 11q – 14} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
q = \frac{1}{2}\\
q = 2,42..
\end{array} \right.
\end{array}\)