Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
8.4.\\
a,\\
\dfrac{{\sqrt {{a^2} + {x^2}} + \sqrt {{a^2} - {x^2}} }}{{\sqrt {{a^2} + {x^2}} - \sqrt {{a^2} - {x^2}} }} - \sqrt {\dfrac{{{a^4}}}{{{x^4}}} - 1} \\
= \dfrac{{{{\left( {\sqrt {{a^2} + {x^2}} + \sqrt {{a^2} - {x^2}} } \right)}^2}}}{{\left( {\sqrt {{a^2} + {x^2}} - \sqrt {{a^2} - {x^2}} } \right)\left( {\sqrt {{a^2} + {x^2}} + \sqrt {{a^2} - {x^2}} } \right)}} - \sqrt {\dfrac{{{a^4} - {x^4}}}{{{x^4}}}} \\
= \dfrac{{\left( {{a^2} + {x^2}} \right) + 2.\sqrt {{a^2} + {x^2}} .\sqrt {{a^2} - {x^2}} + \left( {{a^2} - {x^2}} \right)}}{{\left( {{a^2} + {x^2}} \right) - \left( {{a^2} - {x^2}} \right)}} - \dfrac{{\sqrt {{a^4} - {x^4}} }}{{{x^2}}}\\
= \dfrac{{2{a^2} + 2.\sqrt {{a^4} - {x^4}} }}{{2{x^2}}} - \dfrac{{\sqrt {{a^4} - {x^4}} }}{{{x^2}}}\\
= \dfrac{{{a^2}}}{{{x^2}}} + \dfrac{{\sqrt {{a^4} - {x^4}} }}{{{x^2}}} - \dfrac{{\sqrt {{a^4} - {x^4}} }}{{{x^2}}}\\
= \dfrac{{{a^2}}}{{{x^2}}}\\
b,\\
{\left( {\dfrac{{5 + 2\sqrt 6 }}{{\sqrt 3 + \sqrt 2 }}} \right)^2} - {\left( {\dfrac{{5 - 2\sqrt 6 }}{{\sqrt 3 - \sqrt 2 }}} \right)^2}\\
= {\left( {\dfrac{{3 + 2.\sqrt 3 .\sqrt 2 + 2}}{{\sqrt 3 + \sqrt 2 }}} \right)^2} - {\left( {\dfrac{{3 - 2.\sqrt 3 .\sqrt 2 + 2}}{{\sqrt 3 - \sqrt 2 }}} \right)^2}\\
= {\left( {\dfrac{{{{\left( {\sqrt 3 + \sqrt 2 } \right)}^2}}}{{\sqrt 3 + \sqrt 2 }}} \right)^2} - {\left( {\dfrac{{{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}}}{{\sqrt 3 - \sqrt 2 }}} \right)^2}\\
= {\left( {\sqrt 3 + \sqrt 2 } \right)^2} - {\left( {\sqrt 3 - \sqrt 2 } \right)^2}\\
= 5 + 2\sqrt 6 - \left( {5 - 2\sqrt 6 } \right)\\
= 4\sqrt 6 \\
c,\\
\left( {\dfrac{{x\sqrt x - y\sqrt y }}{{\sqrt x - \sqrt y }} + \sqrt {xy} } \right):{\left( {\sqrt x + \sqrt y } \right)^2}\\
= \left( {\dfrac{{\left( {\sqrt x - \sqrt y } \right)\left( {x + \sqrt {xy} + y} \right)}}{{\sqrt x - \sqrt y }} + \sqrt {xy} } \right):{\left( {\sqrt x + \sqrt y } \right)^2}\\
= \left( {x + \sqrt {xy} + y + \sqrt {xy} } \right):{\left( {\sqrt x + \sqrt y } \right)^2}\\
= {\left( {\sqrt x + \sqrt y } \right)^2}:{\left( {\sqrt x + \sqrt y } \right)^2}\\
= 1\\
8.5\\
a,\\
A = \dfrac{{\sqrt x + 1}}{{\sqrt x - 2}} + \dfrac{{2\sqrt x }}{{\sqrt x + 2}} + \dfrac{{2 + 5\sqrt x }}{{4 - x}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 2}} + \dfrac{{2\sqrt x }}{{\sqrt x + 2}} - \dfrac{{2 + 5\sqrt x }}{{x - 4}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 2}} + \dfrac{{2\sqrt x }}{{\sqrt x + 2}} - \dfrac{{2 + 5\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x + 2} \right) + 2\sqrt x \left( {\sqrt x - 2} \right) - \left( {2 + 5\sqrt x } \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{x + 3\sqrt x + 2 + 2x - 4\sqrt x - 2 - 5\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{3x - 6\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{3\sqrt x \left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{3\sqrt x }}{{\sqrt x + 2}}\\
b,\\
A = 2 \Leftrightarrow \dfrac{{3\sqrt x }}{{\sqrt x + 2}} = 2\\
\Leftrightarrow 3\sqrt x = 2\sqrt x + 4\\
\Leftrightarrow \sqrt x = 4\\
\Leftrightarrow x = 16\,\,\,\,\,\,\,\left( {t/m} \right)
\end{array}\)
