Giải phương trình ; 6x-8=(3x-1) căn(x ^2 -x-1) 19/07/2021 Bởi Samantha Giải phương trình ; 6x-8=(3x-1) căn(x ^2 -x-1)
Đáp án: \[\left[ \begin{array}{l}x = \frac{{ – 1 + 2\sqrt {10} }}{3}\\x = \frac{{ – 1 – 2\sqrt {10} }}{3}\end{array} \right.\] Giải thích các bước giải: ĐK: \({x^2} – x – 1 \ge 0 \Leftrightarrow \left[ \begin{array}{l}x \ge \frac{{1 + \sqrt 5 }}{2}\\x \le \frac{{1 – \sqrt 5 }}{2}\end{array} \right.\) Ta có: \(\begin{array}{l}6x – 8 = \left( {3x – 1} \right)\sqrt {{x^2} – x – 1} \\ \Leftrightarrow {\left( {6x – 8} \right)^2} = {\left( {3x – 1} \right)^2}\left( {{x^2} – x – 1} \right)\\ \Leftrightarrow 36{x^2} – 96x + 64 = \left( {9{x^2} – 6x + 1} \right)\left( {{x^2} – x – 1} \right)\\ \Leftrightarrow 36{x^2} – 96x + 64 = 9{x^4} – 9{x^3} – 9{x^2} – 6{x^3} + 6{x^2} + 6x + {x^2} – x – 1\\ \Leftrightarrow 36{x^2} – 96x + 64 = 9{x^4} – 15{x^3} – 2{x^2} + 5x – 1\\ \Leftrightarrow 9{x^4} – 15{x^3} – 38{x^2} + 101x – 65 = 0\\ \Leftrightarrow \left( {9{x^4} + 6{x^3} – 39{x^2}} \right) – \left( {21{x^3} + 14{x^2} – 91x} \right) + \left( {15{x^2} + 10x – 65} \right) = 0\\ \Leftrightarrow 3{x^2}\left( {3{x^2} + 2x – 13} \right) – 7x\left( {3{x^2} + 2x – 13} \right) + 5\left( {3{x^2} + 2x – 13} \right) = 0\\ \Leftrightarrow \left( {3{x^2} + 2x – 13} \right)\left( {3{x^2} – 7x + 5} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}3{x^2} + 2x – 13 = 0\\3{x^2} – 7x + 5 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{{ – 1 + 2\sqrt {10} }}{3}\\x = \frac{{ – 1 – 2\sqrt {10} }}{3}\end{array} \right.\left( {t/m} \right)\end{array}\) Bình luận
Đáp án:
\[\left[ \begin{array}{l}
x = \frac{{ – 1 + 2\sqrt {10} }}{3}\\
x = \frac{{ – 1 – 2\sqrt {10} }}{3}
\end{array} \right.\]
Giải thích các bước giải:
ĐK: \({x^2} – x – 1 \ge 0 \Leftrightarrow \left[ \begin{array}{l}
x \ge \frac{{1 + \sqrt 5 }}{2}\\
x \le \frac{{1 – \sqrt 5 }}{2}
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
6x – 8 = \left( {3x – 1} \right)\sqrt {{x^2} – x – 1} \\
\Leftrightarrow {\left( {6x – 8} \right)^2} = {\left( {3x – 1} \right)^2}\left( {{x^2} – x – 1} \right)\\
\Leftrightarrow 36{x^2} – 96x + 64 = \left( {9{x^2} – 6x + 1} \right)\left( {{x^2} – x – 1} \right)\\
\Leftrightarrow 36{x^2} – 96x + 64 = 9{x^4} – 9{x^3} – 9{x^2} – 6{x^3} + 6{x^2} + 6x + {x^2} – x – 1\\
\Leftrightarrow 36{x^2} – 96x + 64 = 9{x^4} – 15{x^3} – 2{x^2} + 5x – 1\\
\Leftrightarrow 9{x^4} – 15{x^3} – 38{x^2} + 101x – 65 = 0\\
\Leftrightarrow \left( {9{x^4} + 6{x^3} – 39{x^2}} \right) – \left( {21{x^3} + 14{x^2} – 91x} \right) + \left( {15{x^2} + 10x – 65} \right) = 0\\
\Leftrightarrow 3{x^2}\left( {3{x^2} + 2x – 13} \right) – 7x\left( {3{x^2} + 2x – 13} \right) + 5\left( {3{x^2} + 2x – 13} \right) = 0\\
\Leftrightarrow \left( {3{x^2} + 2x – 13} \right)\left( {3{x^2} – 7x + 5} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
3{x^2} + 2x – 13 = 0\\
3{x^2} – 7x + 5 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{{ – 1 + 2\sqrt {10} }}{3}\\
x = \frac{{ – 1 – 2\sqrt {10} }}{3}
\end{array} \right.\left( {t/m} \right)
\end{array}\)