Giải phương trình
a) \(5x - \sqrt {{{(2x - 1)}^2}} = 2\);                    b) \(\sqrt {x + 2\sqrt {x - 1} } = x\).

a) Điều kiện : \(x > 0;x \ne 9\). Khi đó ta có

\(P = \frac{{x + 3 + \sqrt x  - 3}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}} \cdot \frac{{\sqrt x  - 3}}{{\sqrt x }} = \frac{{\sqrt x \left( {\sqrt x  + 1} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}} \cdot \frac{{\sqrt x  - 3}}{{\sqrt x }} = \frac{{\sqrt x  + 1}}{{\sqrt x  + 3}}\)
b) Xét hiệu\(P - \frac{1}{3} = \frac{{\sqrt x  + 1}}{{\sqrt x  + 3}} - \frac{1}{3} = \frac{{3\sqrt x  + 3 - \sqrt x  - 3}}{{3\left( {\sqrt x  + 3} \right)}} = \frac{{2\sqrt x }}{{3\left( {\sqrt x  + 3} \right)}} > 0({\rm{\;v\`i \;}}x > 0){\rm{.\;}}\)

a) Điều kiện : \(x \ge 0;x \ne 9\). Khi đó ta có

\(P = \frac{{2\sqrt x \left( {\sqrt x  - 3} \right) + \left( {\sqrt x  + 1} \right)\left( {\sqrt x  + 3} \right) - 3 + 11\sqrt x }}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}} = \frac{{2x - 6\sqrt x  + x + 4\sqrt x  + 3 - 3 + 11\sqrt x }}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}}\)

\( = \frac{{3x + 9\sqrt x }}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}} = \frac{{3\sqrt x \left( {\sqrt x  + 3} \right)}}{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right)}} = \frac{{3\sqrt x }}{{\sqrt x  - 3}}\)

b) Ta có \(x = \frac{{7 + 4\sqrt 3 }}{4} = {\left( {\frac{{2 + \sqrt 3 }}{2}} \right)^2} \Rightarrow \sqrt x  = \frac{{2 + \sqrt 3 }}{2}\).
Do đó \({\rm{P}} = \frac{{3 \cdot \frac{{2 + \sqrt 3 }}{2}}}{{\frac{{2 + \sqrt 3 }}{2} - 3}} = \frac{{6 + 3\sqrt 3 }}{2} \cdot \frac{2}{{\sqrt 3  - 4}}\)\( = \frac{{6 + 3\sqrt 3 }}{{\sqrt 3  - 4}} = \frac{{\left( {6 + 3\sqrt 3 } \right)\left( {\sqrt 3  + 4} \right)}}{{\left( {\sqrt 3  - 4} \right)\left( {\sqrt 3  + 4} \right)}}\; = \frac{{6\sqrt 3  + 24 + 9 + 12\sqrt 3 }}{{3 - 16}} = \frac{{ - \left( {33 + 18\sqrt 3 } \right)}}{{13}}\begin{array}{*{20}{r}}{}&\;\\{}&.\end{array}\)