giải pt : $x^{2}$ = (x-1).(3x-2) b) thu gọn ;A=( √3 +1).( √$\frac{14-6√3}{5+√3}$

a)

$\begin{array}{l} {x^2} = \left( {x – 1} \right)\left( {3x – 2} \right)\\  \Leftrightarrow {x^2} = 3{x^2} – 5x + 2\\  \Leftrightarrow 2{x^2} – 5x + 2 = 0\\  \Leftrightarrow \left( {x – 2} \right)\left( {2x – 1} \right) = 0\\  \Leftrightarrow \left[ \begin{array}{l} x = 2\\ x = \dfrac{1}{2} \end{array} \right. \Rightarrow S = \left\{ {2;\dfrac{1}{2}} \right\} \end{array}$  

b)

Thay $\sqrt{3}$ thành $\sqrt 5$
$\begin{array}{l} A = \left( {\sqrt 3  + 1} \right)\dfrac{{\sqrt {14 – 6\sqrt 5 } }}{{5 + \sqrt 3 }}\\ A = \left( {\sqrt 3  + 1} \right).\dfrac{{\sqrt {{{\left( {3 – \sqrt 5 } \right)}^2}} }}{{5 + \sqrt 3 }}\\ A = \left( {\sqrt 3  + 1} \right).\dfrac{{3 – \sqrt 5 }}{{5 + \sqrt 3 }}\\ A = \left( {\sqrt 3  + 1} \right).\dfrac{{\left( {3 – \sqrt 5 } \right)\left( {5 – \sqrt 3 } \right)}}{{25 – 3}}\\ A = \dfrac{{\left( {\sqrt 3  + 1} \right)\left( {15 – 3\sqrt 3  – 5\sqrt 5  + \sqrt {15} } \right)}}{{22}}\\ A = \dfrac{{15\sqrt 3  – 27 – 5\sqrt {15}  + 3\sqrt 5  + 15 – 27 – 5\sqrt 5  + \sqrt {15} }}{{22}}\\ A = \dfrac{{15\sqrt 3  – 39 – 4\sqrt {15}  – 2\sqrt 5 }}{{22}} \end{array}$