Giải phương trình \({x^2} + \frac{{{x^2}}}{{{{\left( {x + 1} \right)}^2}}} = 1\).
A.\(\left[ \begin{array}{l}{x_1} = \frac{{\sqrt 2 + 1 + \sqrt {2\sqrt 2 - 1} }}{2}\\{x_2} = \frac{{\sqrt 2 + 1 - \sqrt {2\sqrt 2 - 1} }}{2}\end{array} \right.\)
B.\(\left[ \begin{array}{l}{x_1} = \frac{{\sqrt 2 - 1 + \sqrt {2\sqrt 2 - 1} }}{2}\\{x_2} = \frac{{\sqrt 2 - 1 - \sqrt {2\sqrt 2 - 1} }}{2}\end{array} \right.\)
C.\(\left[ \begin{array}{l}{x_1} = \frac{{\sqrt 2 - 1 + \sqrt {2\sqrt 2 + 1} }}{2}\\{x_2} = \frac{{\sqrt 2 - 1 - \sqrt {2\sqrt 2 + 1} }}{2}\end{array} \right.\)
D.\(\left[ \begin{array}{l}{x_1} = \frac{{\sqrt 2 + 1 + \sqrt {2\sqrt 2 + 1} }}{2}\\{x_2} = \frac{{\sqrt 2 + 1 - \sqrt {2\sqrt 2 + 1} }}{2}\end{array} \right.\)
A.\(\left[ \begin{array}{l}{x_1} = \frac{{\sqrt 2 + 1 + \sqrt {2\sqrt 2 - 1} }}{2}\\{x_2} = \frac{{\sqrt 2 + 1 - \sqrt {2\sqrt 2 - 1} }}{2}\end{array} \right.\)
B.\(\left[ \begin{array}{l}{x_1} = \frac{{\sqrt 2 - 1 + \sqrt {2\sqrt 2 - 1} }}{2}\\{x_2} = \frac{{\sqrt 2 - 1 - \sqrt {2\sqrt 2 - 1} }}{2}\end{array} \right.\)
C.\(\left[ \begin{array}{l}{x_1} = \frac{{\sqrt 2 - 1 + \sqrt {2\sqrt 2 + 1} }}{2}\\{x_2} = \frac{{\sqrt 2 - 1 - \sqrt {2\sqrt 2 + 1} }}{2}\end{array} \right.\)
D.\(\left[ \begin{array}{l}{x_1} = \frac{{\sqrt 2 + 1 + \sqrt {2\sqrt 2 + 1} }}{2}\\{x_2} = \frac{{\sqrt 2 + 1 - \sqrt {2\sqrt 2 + 1} }}{2}\end{array} \right.\)
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