a) \(A = 1:\left( {\frac{{x + 2\sqrt x - 2}}{{x\sqrt x + 1}} - \frac{{\sqrt x - 1}}{{x - \sqrt x + 1}} + \frac{1}{{\sqrt x + 1}}} \right)\)

\( = 1:\left( {\frac{{x + 2\sqrt x  - 2}}{{x\sqrt x + 1}} - \frac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}} + \frac{{x - \sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}} \right)\)

\( = 1:\left( {\frac{{x + 2\sqrt x - 2}}{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}} + \frac{{ - x + 1 + x - \sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}} \right)\)

\( = 1:\left( {\frac{{x + 2\sqrt x - 2 - \sqrt x + 2}}{{x\sqrt x + 1}}} \right)\)

\( = 1:\frac{{x + \sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}} = \frac{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x  + 1} \right)}} = \frac{{x - \sqrt x + 1}}{{\sqrt x }}\)

b) \(x = 7 - 4\sqrt 3 = 4 - 2.2.\sqrt 3 + 3 = {\left( {2 - \sqrt 3 } \right)^2}\)

Thay vào A ta được:

\(A = \frac{{7 - 4\sqrt 3 - \sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} + 1}}{{\sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} }} = \frac{{7 - 4\sqrt 3 - \left( {2 - \sqrt 3 } \right) + 1}}{{2 - \sqrt 3 }}\)

\( = \frac{{6 - 3\sqrt 3 }}{{2 - \sqrt 3 }} = 3\).