Thực hiện phép tính \(\sqrt {\frac{{3 - 2\sqrt 2 }}{{17 - 12\sqrt 2 }}} - \sqrt {\frac{{3 + 2\sqrt 2 }}{{17 + 12\sqrt 2 }}} \).
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Với các biểu thức A, B, C mà \(A \ge 0,A \ne {B^2}\) ta có \(\frac{C}{{\sqrt A - B}} = \frac{{C\left( {\sqrt A + B} \right)}}{{A - {B^2}}}\), \(\frac{C}{{\sqrt A + B}} = \frac{{C\left( {\sqrt A - B} \right)}}{{A - {B^2}}}\)
\(\sqrt {\frac{{3 - 2\sqrt 2 }}{{17 - 12\sqrt 2 }}} - \sqrt {\frac{{3 + 2\sqrt 2 }}{{17 + 12\sqrt 2 }}} \\ = \sqrt {\frac{{3 - 2\sqrt 2 }}{{{{\left( {2\sqrt 2 } \right)}^2} - 2.2\sqrt 2 .3 + {3^2}}}} - \sqrt {\frac{{3 + 2\sqrt 2 }}{{{{\left( {2\sqrt 2 } \right)}^2} + 2.2\sqrt 2 .3 + {3^2}}}} \\ = \sqrt {\frac{{3 - 2\sqrt 2 }}{{{{\left( {2\sqrt 2 - 3} \right)}^2}}}} - \sqrt {\frac{{3 + 2\sqrt 2 }}{{{{\left( {2\sqrt 2 + 3} \right)}^2}}}} \\ = \sqrt {\frac{1}{{3 - 2\sqrt 2 }}} - \sqrt {\frac{1}{{3 + 2\sqrt 2 }}} \\ = \sqrt {\frac{{3 + 2\sqrt 2 }}{{\left( {3 - 2\sqrt 2 } \right)\left( {3 + 2\sqrt 2 } \right)}}} - \sqrt {\frac{{3 - 2\sqrt 2 }}{{\left( {3 + 2\sqrt 2 } \right)\left( {3 - 2\sqrt 2 } \right)}}} \\ = \sqrt {{{\left( {1 + \sqrt 2 } \right)}^2}} - \sqrt {{{\left( { - 1 + \sqrt 2 } \right)}^2}} \)\( = 1 + \sqrt 2 - \sqrt 2 + 1 = 2\)