Question

giúp mình vs mọi người ạ

$\begin{array}{l} A = \dfrac{a}{{\sqrt {ab} + b}} + \dfrac{b}{{\sqrt {ab} – a}} – \dfrac{{a + b}}{{\sqrt {ab} }}\left( {a > 0;b > 0;a \ne b} \right)\\ = \dfrac{a}{{\sqrt b \left( {\sqrt a + \sqrt b } \right)}} + \dfrac{b}{{\sqrt a \left( {\sqrt b – \sqrt a } \right)}} – \dfrac{{a + b}}{{\sqrt {ab} }}\\ = \dfrac{{a.\sqrt a .\left( {\sqrt b – \sqrt a } \right) + b\sqrt b \left( {\sqrt a + \sqrt b } \right) – \left( {a + b} \right).\left( {b – a} \right)}}{{\sqrt {ab} \left( {b – a} \right)}}\\ = \dfrac{{a\sqrt {ab} – {a^2} + b\sqrt {ab} + {b^2} – {b^2} + {a^2}}}{{\sqrt {ab} \left( {b – a} \right)}}\\ = \dfrac{{\sqrt {ab} \left( {a + b} \right)}}{{\sqrt {ab} \left( {b – a} \right)}}\\ = \dfrac{{a + b}}{{b – a}}\\ Khi:\\ a = \sqrt {4 + 2\sqrt 3 } \left( {tmdk} \right)\\ = \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} = \sqrt 3 + 1\\ b = \sqrt {4 – 2\sqrt 3 } \left( {tmdk} \right)\\ = \sqrt 3 – 1\\ \Rightarrow A = \dfrac{{a + b}}{{b – a}}\\ = \dfrac{{\sqrt 3 + 1 + \sqrt 3 – 1}}{{\sqrt 3 – 1 – \sqrt 3 – 1}}\\ = \dfrac{{2\sqrt 3 }}{{ – 2}} = – \sqrt 3 \\ b)\dfrac{a}{b} = \dfrac{{a + 1}}{{b + 5}}\\ \Rightarrow a.\left( {b + 5} \right) = b.\left( {a + 1} \right)\\ \Rightarrow ab + 5a = ab + b\\ \Rightarrow 5a = b\\ \Rightarrow A = \dfrac{{a + b}}{{b – a}}\\ = \dfrac{{a + 5a}}{{5a – a}} = \dfrac{{6a}}{{4a}} = \dfrac{3}{2}{\rm{ = const}} \end{array}$