giải phương trình X+ √2-X ² + X × √2-X ²=3

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giải phương trình X+ √2-X ² + X × √2-X ²=3

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4 tháng 2021-08-14T19:57:08+00:00 1 Answers 7 views 0

$\begin{array}{l} x + \sqrt {2 – {x^2}} + x\sqrt {2 – {x^2}} = 3\left( {dkxd:2 – {x^2} \ge 0 \Rightarrow – \sqrt 2 \le x \le \sqrt 2 } \right)\\ Đặt:x + \sqrt {2 – {x^2}} = t\\ \Rightarrow {x^2} + 2x\sqrt {2 – {x^2}} + 2 – {x^2} = {t^2}\\ \Rightarrow 2x\sqrt {2 – {x^2}} = {t^2} – 2\\ \Rightarrow x\sqrt {2 – {x^2}} = \frac{{{t^2} – 2}}{2}\\ Pt \Rightarrow t + \frac{{{t^2} – 2}}{2} = 3\\ \Rightarrow {t^2} + 2t – 2 – 6 = 0\\ \Rightarrow {t^2} + 4t – 2t – 8 = 0\\ \Rightarrow \left[ \begin{array}{l} t = – 4\\ t = 2 \end{array} \right.\\ \Rightarrow \left[ \begin{array}{l} x\sqrt {2 – {x^2}} = 7\\ x\sqrt {2 – {x^2}} = 1 \end{array} \right. \Rightarrow \left[ \begin{array}{l} {x^2}\left( {2 – {x^2}} \right) = 49\\ {x^2}\left( {2 – {x^2}} \right) = 1 \end{array} \right.\\ \Rightarrow \left[ \begin{array}{l} {x^4} – 2{x^2} + 49 = 0\left( {loai} \right)\\ {x^4} – 2{x^2} + 1 = 0 \end{array} \right.\\ \Rightarrow {x^2} = 1\\ \Rightarrow x = \pm 1\left( {tmdk} \right) \end{array}$