giải phương trình: 5/x^2-4x+5 – (x^2+4x-1)=0 02/09/2021 Bởi Mary giải phương trình: 5/x^2-4x+5 – (x^2+4x-1)=0
Đáp án: $\begin{array}{l}Đặt:{x^2} – 4x + 5 = t\\ \Rightarrow t = {\left( {x – 2} \right)^2} + 1 \ge 1 > 0\\ \Rightarrow {x^2} + 4x – 1 = {x^2} – 4x + 5 – 6 = t – 6\\Pt:\dfrac{5}{{{x^2} – 4x + 5}} – \left( {{x^2} – 4x – 1} \right) = 0\\ \Rightarrow \dfrac{5}{t} – \left( {t – 6} \right) = 0\\ \Rightarrow 5 – {t^2} + 6t = 0\left( {do:t > 0} \right)\\ \Rightarrow {t^2} – 6t – 5 = 0\\ \Rightarrow {t^2} – 6t + 9 – 14 = 0\\ \Rightarrow {\left( {t – 3} \right)^2} = 14\\ \Rightarrow t = 3 + \sqrt {14} \left( {do:t > 0} \right)\\ \Rightarrow {x^2} – 4x + 5 = 3 + \sqrt {14} \\ \Rightarrow {x^2} – 4x + 4 = 2 + \sqrt {14} \\ \Rightarrow {\left( {x – 2} \right)^2} = 2 + \sqrt {14} \\ \Rightarrow \left[ \begin{array}{l}x = 2 + \sqrt {2 + \sqrt {14} } \\x = 2 – \sqrt {2 + \sqrt {14} } \end{array} \right.\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
Đặt:{x^2} – 4x + 5 = t\\
\Rightarrow t = {\left( {x – 2} \right)^2} + 1 \ge 1 > 0\\
\Rightarrow {x^2} + 4x – 1 = {x^2} – 4x + 5 – 6 = t – 6\\
Pt:\dfrac{5}{{{x^2} – 4x + 5}} – \left( {{x^2} – 4x – 1} \right) = 0\\
\Rightarrow \dfrac{5}{t} – \left( {t – 6} \right) = 0\\
\Rightarrow 5 – {t^2} + 6t = 0\left( {do:t > 0} \right)\\
\Rightarrow {t^2} – 6t – 5 = 0\\
\Rightarrow {t^2} – 6t + 9 – 14 = 0\\
\Rightarrow {\left( {t – 3} \right)^2} = 14\\
\Rightarrow t = 3 + \sqrt {14} \left( {do:t > 0} \right)\\
\Rightarrow {x^2} – 4x + 5 = 3 + \sqrt {14} \\
\Rightarrow {x^2} – 4x + 4 = 2 + \sqrt {14} \\
\Rightarrow {\left( {x – 2} \right)^2} = 2 + \sqrt {14} \\
\Rightarrow \left[ \begin{array}{l}
x = 2 + \sqrt {2 + \sqrt {14} } \\
x = 2 – \sqrt {2 + \sqrt {14} }
\end{array} \right.
\end{array}$