Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
DKXD:\,\,\,\, - 1 < x < 1\\
\left( {\dfrac{3}{{\sqrt {x + 1} }} - \sqrt {1 - x} } \right):\left( {\dfrac{3}{{\sqrt {1 - {x^2}} }} + 1} \right)\\
= \dfrac{{3 - \sqrt {1 - x} .\sqrt {x + 1} }}{{\sqrt {x + 1} }}:\dfrac{{3 + \sqrt {1 - {x^2}} }}{{\sqrt {1 - {x^2}} }}\\
= \dfrac{{3 - \sqrt {1 - {x^2}} }}{{\sqrt {x + 1} }}:\dfrac{{3 + \sqrt {1 - {x^2}} }}{{\sqrt {1 - x} .\sqrt {1 + x} }}\\
= \dfrac{{3 - \sqrt {1 - {x^2}} }}{{\sqrt {x + 1} }}.\dfrac{{\sqrt {1 - x} .\sqrt {1 + x} }}{{3 + \sqrt {1 - {x^2}} }}\\
= \dfrac{{\left( {3 - \sqrt {1 - {x^2}} } \right).\sqrt {1 - x} }}{{3 + \sqrt {1 - {x^2}} }}\\
= \dfrac{{{{\left( {3 - \sqrt {1 - {x^2}} } \right)}^2}.\sqrt {1 - x} }}{{\left( {3 + \sqrt {1 - {x^2}} } \right).\left( {3 - \sqrt {1 - {x^2}} } \right)}}\\
= \dfrac{{\left( {9 - 6\sqrt {1 - {x^2}} + 1 - {x^2}} \right).\sqrt {1 - x} }}{{9 - \left( {1 - {x^2}} \right)}}\\
= \dfrac{{\left( {{x^2} - 6\sqrt {1 - {x^2}} + 10} \right).\sqrt {1 - x} }}{{{x^2} + 8}}
\end{array}\)
