Giải phương trình bằng phương pháp đặt t= √P(x) ± √Q(x) √2x-3 + √5-2x – x^2+4x-6=0 01/09/2021 Bởi Eloise Giải phương trình bằng phương pháp đặt t= √P(x) ± √Q(x) √2x-3 + √5-2x – x^2+4x-6=0
Đáp án: \(x=2.\) Giải thích các bước giải: \(\begin{array}{l}\sqrt {2x – 3} + \sqrt {5 – 2x} – {x^2} + 4x – 6 = 0\,\,\,\,\,\left( * \right)\\DK:\,\,\,\frac{3}{2} \le x \le \frac{5}{2}\\Dat\,\,t = \sqrt {2x – 3} + \sqrt {5 – 2x} \,\,\left( {t \ge 0} \right)\\ \Rightarrow {t^2} = 2x – 3 + 5 – 2x + 2\sqrt {\left( {2x – 3} \right)\left( {5 – 2x} \right)} \\ \Leftrightarrow {t^2} = 2 + 2\sqrt { – 4{x^2} + 16x – 15} \\ \Leftrightarrow {t^2} – 2 = 2\sqrt { – 4\left( {{x^2} – 4x} \right) – 15} \\ \Leftrightarrow {\left( {{t^2} – 2} \right)^2} = 4\left[ { – 4\left( {{x^2} – 4x} \right) – 15} \right]\\ \Leftrightarrow {t^4} – 4{t^2} + 4 = – 16\left( {{x^2} – 4x} \right) – 60\\ \Leftrightarrow – \left( {{x^2} – 4x} \right) = \frac{{{t^4} – 4{t^2} + 64}}{{16}}\\ \Rightarrow \left( * \right) \Leftrightarrow t + \frac{{{t^4} – 4{t^2} + 64}}{{16}} – 6 = 0\\ \Leftrightarrow 16t + {t^4} – 4{t^2} + 64 – 96 = 0\\ \Leftrightarrow {t^4} – 4{t^2} + 16t – 32 = 0\\ \Leftrightarrow {t^2}\left( {{t^2} – 4} \right) + 16\left( {t – 2} \right) = 0\\ \Leftrightarrow {t^2}\left( {t – 2} \right)\left( {t + 2} \right) + 16\left( {t – 2} \right) = 0\\ \Leftrightarrow \left( {t – 2} \right)\left[ {{t^2}\left( {t + 2} \right) + 16} \right] = 0\\ \Leftrightarrow t – 2 = 0\,\,\,\,\,\left( {do\,\,\,{t^2}\left( {t + 2} \right) + 16 > 0\,\,\,\forall t \ge 0} \right)\\ \Leftrightarrow t = 2\,\,\,\left( {tm} \right)\\ \Leftrightarrow – \left( {{x^2} – 4x} \right) = \frac{{{t^4} – 4{t^2} + 64}}{{16}} = 4\\ \Leftrightarrow – {x^2} + 4x – 4 = 0\\ \Leftrightarrow {\left( {x – 2} \right)^2} = 0\\ \Leftrightarrow x – 2 = 0\\ \Leftrightarrow x = 2\,\,\,\left( {tm} \right).\end{array}\) Bình luận
Đáp án:
Đáp án:
\(x=2.\)
Giải thích các bước giải:
\(\begin{array}{l}
\sqrt {2x – 3} + \sqrt {5 – 2x} – {x^2} + 4x – 6 = 0\,\,\,\,\,\left( * \right)\\
DK:\,\,\,\frac{3}{2} \le x \le \frac{5}{2}\\
Dat\,\,t = \sqrt {2x – 3} + \sqrt {5 – 2x} \,\,\left( {t \ge 0} \right)\\
\Rightarrow {t^2} = 2x – 3 + 5 – 2x + 2\sqrt {\left( {2x – 3} \right)\left( {5 – 2x} \right)} \\
\Leftrightarrow {t^2} = 2 + 2\sqrt { – 4{x^2} + 16x – 15} \\
\Leftrightarrow {t^2} – 2 = 2\sqrt { – 4\left( {{x^2} – 4x} \right) – 15} \\
\Leftrightarrow {\left( {{t^2} – 2} \right)^2} = 4\left[ { – 4\left( {{x^2} – 4x} \right) – 15} \right]\\
\Leftrightarrow {t^4} – 4{t^2} + 4 = – 16\left( {{x^2} – 4x} \right) – 60\\
\Leftrightarrow – \left( {{x^2} – 4x} \right) = \frac{{{t^4} – 4{t^2} + 64}}{{16}}\\
\Rightarrow \left( * \right) \Leftrightarrow t + \frac{{{t^4} – 4{t^2} + 64}}{{16}} – 6 = 0\\
\Leftrightarrow 16t + {t^4} – 4{t^2} + 64 – 96 = 0\\
\Leftrightarrow {t^4} – 4{t^2} + 16t – 32 = 0\\
\Leftrightarrow {t^2}\left( {{t^2} – 4} \right) + 16\left( {t – 2} \right) = 0\\
\Leftrightarrow {t^2}\left( {t – 2} \right)\left( {t + 2} \right) + 16\left( {t – 2} \right) = 0\\
\Leftrightarrow \left( {t – 2} \right)\left[ {{t^2}\left( {t + 2} \right) + 16} \right] = 0\\
\Leftrightarrow t – 2 = 0\,\,\,\,\,\left( {do\,\,\,{t^2}\left( {t + 2} \right) + 16 > 0\,\,\,\forall t \ge 0} \right)\\
\Leftrightarrow t = 2\,\,\,\left( {tm} \right)\\
\Leftrightarrow – \left( {{x^2} – 4x} \right) = \frac{{{t^4} – 4{t^2} + 64}}{{16}} = 4\\
\Leftrightarrow – {x^2} + 4x – 4 = 0\\
\Leftrightarrow {\left( {x – 2} \right)^2} = 0\\
\Leftrightarrow x – 2 = 0\\
\Leftrightarrow x = 2\,\,\,\left( {tm} \right).
\end{array}\)