Lời giải: \(Q = \dfrac{a}{{\sqrt {{a^2} - {b^2}} }} - \left( {1 + \dfrac{a}{{\sqrt {{a^2} - {b^2}} }}} \right):\dfrac{b}{{a - \sqrt {{a^2} - {b^2}} }}\)
\(= \dfrac{a}{{\sqrt {{a^2} - {b^2}} }} - \left( {\dfrac{{\sqrt {{a^2} - {b^2}} + a}}{{\sqrt {{a^2} - {b^2}} }}} \right) . \dfrac{{a - \sqrt {{a^2} - {b^2}} }}{b}\)
\(= \dfrac{a}{{\sqrt {{a^2} - {b^2}} }} - \dfrac{{\left( {\sqrt {{a^2} - {b^2}} + a} \right)\left( {a - \sqrt {{a^2} - {b^2}} } \right)}}{{b\sqrt {{a^2} - {b^2}} }}\)
\(= \dfrac{a}{{\sqrt {{a^2} - {b^2}} }} - \dfrac{{{a^2} - {{\left( {\sqrt {{a^2} - {b^2}} } \right)}^2}}}{{b\sqrt {{a^2} - {b^2}} }}\)
\(= \dfrac{a}{{\sqrt {{a^2} - {b^2}} }} - \dfrac{{{a^2} - {a^2} + {b^2}}}{{b\sqrt {{a^2} - {b^2}} }}\)
\(= \dfrac{a}{{\sqrt {{a^2} - {b^2}} }} - \dfrac{{{b^2}}}{{b\sqrt {{a^2} - {b^2}} }}\)
\(= \dfrac{{ab - {b^2}}}{{b\sqrt {{a^2} - {b^2}} }}\)
\(= \dfrac{{b\left( {a - b} \right)}}{{b\sqrt {{a^2} - {b^2}} }}\)
\(= \dfrac{{a - b}}{{\sqrt {{a^2} - {b^2}} }}\)
