a) Cho biểu thức P = ( x + 3 căn bậc hai x + 2 / x căn bậc hai x -8 - 1/ căn bậc hai x -2 ) : 1/ căn bậc hai x

Với \(x > 0;x \ne 4\), ta có:

\[P = \left( {\frac{{x + 3\sqrt x  + 2}}{{x\sqrt x  - 8}} - \frac{1}{{\sqrt x  - 2}}} \right):\frac{1}{{\sqrt x }}\]

\( = \left[ {\frac{{x + 3\sqrt x  + 2}}{{\left( {\sqrt x  - 2} \right)\left( {x + 2\sqrt x  + 4} \right)}} - \frac{{x + 2\sqrt x  + 4}}{{\left( {\sqrt x  - 2} \right)\left( {x + 2\sqrt x  + 4} \right)}}} \right].\sqrt x \)

\( = \frac{{\sqrt x  - 2}}{{\left( {\sqrt x  - 2} \right)\left( {x + 2\sqrt x  + 4} \right)}}.\sqrt x  = \frac{{\sqrt x }}{{x + 2\sqrt x  + 4}}.\)

Ta có

\(x = 14 + 6\sqrt 5  = 9 + 2.3.\sqrt 5  + 5 = {\left( {3 + \sqrt 5 } \right)^2} \Rightarrow \sqrt x  = \sqrt {{{\left( {3 + \sqrt 5 } \right)}^2}}  = \left| {3 + \sqrt 5 } \right| = 3 + \sqrt 5 .\)

Khi đó, ta có: \(P = \frac{{3 + \sqrt 5 }}{{14 + 6\sqrt 5  + 2.\left( {3 + \sqrt 5 } \right) + 4}} = \frac{{3 + \sqrt 5 }}{{24 + 8\sqrt 5 }} = \frac{{3 + \sqrt 5 }}{{8.\left( {3 + \sqrt 5 } \right)}} = \frac{1}{8}.\)

b) \(\sqrt {\frac{{3 - 2\sqrt 2 }}{{{{\left( {3 - 2\sqrt 2 } \right)}^2}}}}  - \sqrt {\frac{{3 + 2\sqrt 2 }}{{{{\left( {3 + 2\sqrt 2 } \right)}^2}}}}  = \sqrt {\frac{1}{{3 - 2\sqrt 2 }}}  - \sqrt {\frac{1}{{3 + 2\sqrt 2 }}} \)

\( = \frac{1}{{\left| {\sqrt 2  - 1} \right|}} - \frac{1}{{\left| {\sqrt 2  + 1} \right|}} = \frac{1}{{\sqrt 2  - 1}} - \frac{1}{{\sqrt 2  + 1}} = 2\) (vì \(\sqrt 2  - 1 > 0\) )