Lời giải chi tiết:

a) Cách 1:

                 \(\begin{array}{l}\sqrt {4{x^2} - 4x + 1}  = \left| {x - 2} \right|\\ \Leftrightarrow \sqrt {{{\left( {2x - 1} \right)}^2}}  = \left| {x - 2} \right|\\ \Leftrightarrow \left| {2x - 1} \right| = \left| {x - 2} \right|\\ \Leftrightarrow \left[ \begin{array}{l}2x - 1 = x - 2\\2x - 1 =  - x + 2\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}2x - x =  - 2 + 1\\2x + x = 2 + 1\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x =  - 1\\3x = 3\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x =  - 1\\x = 1\end{array} \right.\end{array}\)

Cách 2:

                 \(\begin{array}{l}\sqrt {4{x^2} - 4x + 1}  = \left| {x - 2} \right|\\ \Leftrightarrow {\left( {\sqrt {4{x^2} - 4x + 1} } \right)^2} = {\left( {x - 2} \right)^2}\\ \Leftrightarrow 4{x^2} - 4x + 1 = {x^2} - 4x + 4\\ \Leftrightarrow 4{x^2} - {x^2} = 4 - 1\\ \Leftrightarrow 3{x^2} = 3\\ \Leftrightarrow {x^2} = 1\\ \Leftrightarrow x =  \pm 1\end{array}\)

Vậy:  Nghiệm phương trình là: \(S=\left\{ -1;1 \right\}\)

b) \(\begin{array}{l}A = \frac{{5 + \sqrt 5 }}{{\sqrt 5  + 2}} + \frac{{\sqrt 5 }}{{\sqrt 5  - 1}} - \frac{{3\sqrt 5 }}{{3 + \sqrt 5 }}\\A = \frac{{\left( {5 + \sqrt 5 } \right)\left( {\sqrt 5  - 2} \right)}}{{\left( {\sqrt 5  + 2} \right)\left( {\sqrt 5  - 2} \right)}} + \frac{{\sqrt 5 \left( {\sqrt 5  + 1} \right)}}{{\left( {\sqrt 5  - 1} \right)\left( {\sqrt 5  + 1} \right)}} - \frac{{3\sqrt 5 \left( {3 - \sqrt 5 } \right)}}{{\left( {3 + \sqrt 5 } \right)\left( {3 - \sqrt 5 } \right)}}\\A = \frac{{ - 5 + 3\sqrt 5 }}{1} + \frac{{5 + \sqrt 5 }}{4} - \frac{{ - 15 + 9\sqrt 5 }}{4}\\A = \frac{{ - 20 + 12\sqrt 5 }}{4} + \frac{{5 + \sqrt 5 }}{4} - \frac{{ - 15 + 9\sqrt 5 }}{4}\\A = \frac{{ - 20 + 12\sqrt 5  + 5 + \sqrt 5  + 15 - 9\sqrt 5 }}{4}\\A = \frac{{4\sqrt 5 }}{4}\\A = \sqrt 5 \end{array}\)