Bài 1. a. Ta có: \(A = \sqrt {{{2\left( {3 - \sqrt 5 } \right)} \over {{{\left( {3 - \sqrt 5 } \right)}^2}}}} = {1 \over {3 - \sqrt 5 }}\sqrt {6 - 2\sqrt 5 } \)\(\, = {{\sqrt 5 - 1} \over {3 - \sqrt 5 }}\)
b. Ta có: \(B = \sqrt {{{\left( {\sqrt a + 2} \right)\left( {\sqrt a - 2} \right)} \over {2\left( {\sqrt a - 2} \right)}}} = {1 \over 2}\sqrt {2\left( {\sqrt a + 2} \right)} \)\(\,\,\,\,\,\,\,\left( {a \ge 0;a \ne 4} \right)\)
Bài 2. \(VT = {1 \over {\sqrt 3 + \sqrt 2 }} = {{\sqrt 3 - \sqrt 2 } \over {3 - 2}} = \sqrt 3 - \sqrt 2 \)\(\, = VP\,\,\,\left( {đpcm} \right)\)
Bài 3. Ta có: \({{3\sqrt 7 + 5\sqrt 2 } \over {\sqrt 5 }} = {{\left( {3\sqrt 7 + 5\sqrt 2 } \right)\sqrt 5 } \over 5} \)\(\,= {3 \over 5}\sqrt {35} + \sqrt {10} < \sqrt {35} + \sqrt {10} \)
Vậy : \({{3\sqrt 7 + 5\sqrt 2 } \over {\sqrt 5 }} < \sqrt {35} + \sqrt {10} \)
