giai pt x^2 + 6x +1= (2x+1)can bac hai x^2+2x+3 25/07/2021 Bởi Iris giai pt x^2 + 6x +1= (2x+1)can bac hai x^2+2x+3
Đáp án: $S = \{ – 1 \pm \sqrt 2 ;\frac{{3 + \sqrt {15} }}{3}\} $ Giải thích các bước giải: ${x^2} + 6x + 1 = (2x + 1)\sqrt {{x^2} + 2x + 3} (1)$ $\begin{array}{l}u = \sqrt {{x^2} + 2x + 3} \\ = > {u^2} = {x^2} + 2x + 3\\(1) = > {u^2} – (2x + 1).u + 4x – 2 = 0\\ < = > {u^2} – (2x – 1 + 2) + 2(2x – 1) = 0\\ < = > [_{u = 2x – 1}^{u = 2}\\ + u = 2 = > \sqrt {{x^2} + 2x + 3} = 2\\ < = > {x^2} + 2x + 3 = 4\\ < = > {x^2} + 2x – 1 = 0\\ < = > [_{x = – 1 – \sqrt 2 }^{x = – 1 + \sqrt 2 }\\ + u = 2x – 1 = > \sqrt {{x^2} + 2x + 3} = 2x – 1\\ < = > \{ _{{x^2} + 2x + 3 = 4{x^2} – 4x + 1}^{2x – 1 \ge 0}\\ < = > \{ _{3{x^2} – 6x – 2 = 0}^{x \ge \frac{1}{2}}\\ < = > \{ _{[_{x = \frac{{3 – \sqrt {15} }}{3}(loại)}^{x = \frac{{3 + \sqrt {15} }}{3}(nhận)}}^{x \ge \frac{1}{2}}\\ = > S = \{ – 1 \pm \sqrt 2 ;\frac{{3 + \sqrt {15} }}{3}\} \end{array}$ Bình luận
Đáp án:
$S = \{ – 1 \pm \sqrt 2 ;\frac{{3 + \sqrt {15} }}{3}\} $
Giải thích các bước giải:
${x^2} + 6x + 1 = (2x + 1)\sqrt {{x^2} + 2x + 3} (1)$
$\begin{array}{l}
u = \sqrt {{x^2} + 2x + 3} \\
= > {u^2} = {x^2} + 2x + 3\\
(1) = > {u^2} – (2x + 1).u + 4x – 2 = 0\\
< = > {u^2} – (2x – 1 + 2) + 2(2x – 1) = 0\\
< = > [_{u = 2x – 1}^{u = 2}\\
+ u = 2 = > \sqrt {{x^2} + 2x + 3} = 2\\
< = > {x^2} + 2x + 3 = 4\\
< = > {x^2} + 2x – 1 = 0\\
< = > [_{x = – 1 – \sqrt 2 }^{x = – 1 + \sqrt 2 }\\
+ u = 2x – 1 = > \sqrt {{x^2} + 2x + 3} = 2x – 1\\
< = > \{ _{{x^2} + 2x + 3 = 4{x^2} – 4x + 1}^{2x – 1 \ge 0}\\
< = > \{ _{3{x^2} – 6x – 2 = 0}^{x \ge \frac{1}{2}}\\
< = > \{ _{[_{x = \frac{{3 – \sqrt {15} }}{3}(loại)}^{x = \frac{{3 + \sqrt {15} }}{3}(nhận)}}^{x \ge \frac{1}{2}}\\
= > S = \{ – 1 \pm \sqrt 2 ;\frac{{3 + \sqrt {15} }}{3}\}
\end{array}$