Đáp án:
2) a. \(\left[ \begin{array}{l}
x = 1 + 2\sqrt 2 \\
x = 1 - 2\sqrt 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)DK: - 1 < x < 1\\
F = \left( {\dfrac{{3 + \sqrt {1 - {a^2}} }}{{\sqrt {1 + a} }}} \right):\dfrac{3}{{\sqrt {1 - {a^2}} }}\\
= \dfrac{{3 + \sqrt {1 - {a^2}} }}{{\sqrt {1 + a} }}.\dfrac{{\sqrt {1 + a} .\sqrt {1 - a} }}{3}\\
= \dfrac{{\left( {3 + \sqrt {1 - {a^2}} } \right)\left( {\sqrt {1 - a} } \right)}}{3}\\
Thay:a = \dfrac{{\sqrt 3 }}{{2 + \sqrt 3 }} = \dfrac{{\sqrt 3 \left( {2 - \sqrt 3 } \right)}}{{4 - 3}}\\
= \dfrac{{ - 3 + 2\sqrt 3 }}{1} = 2\sqrt 3 - 3\\
\to F = \dfrac{{\left( {3 + \sqrt {1 - {{\left( {2\sqrt 3 - 3} \right)}^2}} } \right)\left( {\sqrt {1 - \left( {2\sqrt 3 - 3} \right)} } \right)}}{3}\\
= \dfrac{{\left[ {3 + \sqrt {1 - \left( {21 - 12\sqrt 3 } \right)} } \right]\left( {\sqrt {1 - \left( {2\sqrt 3 - 3} \right)} } \right)}}{3}\\
= \dfrac{{\left( {3 + \sqrt {12\sqrt 3 - 20} } \right)\left( {\sqrt {4 - 2\sqrt 3 } } \right)}}{3}\\
= \dfrac{{\left( {3 + \sqrt {12\sqrt 3 - 20} } \right)\left( {\sqrt {{{\left( {\sqrt 3 } \right)}^2} - 2.\sqrt 3 .1 + 1} } \right)}}{3}\\
= \dfrac{{\left( {3 + \sqrt {12\sqrt 3 - 20} } \right)\sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} }}{3}\\
= \dfrac{{\left( {3 + \sqrt {12\sqrt 3 - 20} } \right)\left( {\sqrt 3 - 1} \right)}}{3}
\end{array}\)
\(\begin{array}{l}
2)a.2x - {x^2} + \sqrt {6{x^2} - 12x + 7} = 0\\
\to \sqrt {6{x^2} - 12x + 7} = {x^2} - 2x\left( 1 \right)\\
Đặt:\sqrt {6{x^2} - 12x + 7} = t\left( {t \ge 0} \right)\\
\to 6{x^2} - 12x + 7 = {t^2}\\
\to 6{x^2} - 12x = {t^2} - 7\\
\to {x^2} - 2x = \dfrac{{{t^2} - 7}}{6}\\
Pt\left( 1 \right) \to t = \dfrac{{{t^2} - 7}}{6}\\
\to {t^2} - 6t - 7 = 0\\
\to {t^2} + t - 7x - 7 = 0\\
\to t\left( {t + 1} \right) - 7\left( {t + 1} \right) = 0\\
\to \left[ \begin{array}{l}
t = - 1\left( l \right)\\
t = 7
\end{array} \right.\\
\to \sqrt {6{x^2} - 12x + 7} = 7\\
\to 6{x^2} - 12x + 7 = 49\\
\to \left[ \begin{array}{l}
x = 1 + 2\sqrt 2 \\
x = 1 - 2\sqrt 2
\end{array} \right.
\end{array}\)
