a) Ta có:
\( \displaystyle{{\sqrt a + \sqrt b } \over {\sqrt a - \sqrt b }} + {{\sqrt a - \sqrt b } \over {\sqrt a + \sqrt b }} \) \( \displaystyle= {{{{\left( {\sqrt a + \sqrt b } \right)}^2} + {{\left( {\sqrt a - \sqrt b } \right)}^2}} \over {\left( {\sqrt a + \sqrt b } \right)\left( {\sqrt a - \sqrt b } \right)}}\)
\( \displaystyle = {{a + 2\sqrt {ab} + b + a - 2\sqrt {ab} + b} \over {a - b}}\)
\( \displaystyle = {{2(a + b)} \over {a - b}}\) (với \(a \ge 0,b \ge 0\) và \( a \ne b\))
b) Ta có: \( \displaystyle{{a - b} \over {\sqrt a - \sqrt b }} -{{\sqrt {a^3} - \sqrt {{b^3}} } \over {a - b}}\)
\( \displaystyle = {{(a - b)(\sqrt a + \sqrt {b)} } \over {{{\left( {\sqrt a } \right)}^2} - {{\left( {\sqrt b } \right)}^2}}} - {{a\sqrt a - b\sqrt b } \over {a - b}}\)
\( \displaystyle = {{a\sqrt a + a\sqrt b - b\sqrt a - b\sqrt b } \over {a - b}} - {{a\sqrt a - b\sqrt b } \over {a - b}}\)
\( \displaystyle = {{a\sqrt a + a\sqrt b - b\sqrt a - b\sqrt b - a\sqrt a + b\sqrt b } \over {a - b}}\)
\( \displaystyle = {{a\sqrt b - b\sqrt a } \over {a - b}}\) (với \(a \ge 0,b \ge 0\) và \(a \ne b\))