Lời giải:
\(A = \frac{{x\sqrt x - 3 - 2{{\left( {\sqrt x - 3} \right)}^2} - \left( {\sqrt x + 3} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 1} \right)}}\)
\( = \frac{{x\sqrt x - 3 - 2x + 12\sqrt x - 18 - x - 4\sqrt x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 1} \right)}}\)
\(\begin{array}{l} = \frac{{x\sqrt x - 3x + 8\sqrt x - 24}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 1} \right)}}\\ = \frac{{x\left( {\sqrt x - 3} \right) + 8\left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 1} \right)}}\\ = \frac{{\left( {\sqrt x - 3} \right)\left( {x + 8} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 1} \right)}} = \frac{{x + 8}}{{\sqrt x + 1}}\end{array}\)
Ta có: \(x = 4 - 2\sqrt 3 \) thỏa mãn ĐKXĐ:
\(x = 4 - 2\sqrt 3 = {\left( {\sqrt 3 - 1} \right)^2} \Rightarrow \sqrt x = \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} = \sqrt 3 - 1\) (Do \(\sqrt 3 - 1 > 0\))
Thay \(x = 4 - 2\sqrt 3 \) và \(\sqrt x = \sqrt 3 - 1\) vào biểu thức A ta có:
\(A = \frac{{4 - 2\sqrt 3 + 8}}{{\sqrt 3 - 1 + 1}} = \frac{{12 - 2\sqrt 3 }}{{\sqrt 3 }} = 4\sqrt 3 - 2\)
Vậy khi x = 4 – \(2\sqrt 3 \) thì \(A = 4\sqrt 3 - 2\).