Đáp án:
$\begin{array}{l}
b)\left( {\dfrac{{\sqrt {14} - \sqrt 7 }}{{1 - \sqrt 2 }} + \dfrac{{\sqrt {15} - \sqrt 5 }}{{1 - \sqrt 3 }}} \right):\dfrac{1}{{\sqrt 7 - \sqrt 5 }}\\
= \left( {\dfrac{{ - \sqrt 7 \left( {1 - \sqrt 2 } \right)}}{{1 - \sqrt 2 }} - \dfrac{{\sqrt 5 \left( {1 - \sqrt 3 } \right)}}{{1 - \sqrt 3 }}} \right).\left( {\sqrt 7 - \sqrt 5 } \right)\\
= \left( { - \sqrt 7 - \sqrt 5 } \right)\left( {\sqrt 7 - \sqrt 5 } \right)\\
= - \left( {7 - 5} \right)\\
= - 2\\
c)\dfrac{{\left( {3 + \sqrt 5 } \right)\sqrt {3 - \sqrt 5 } }}{{\sqrt {3 + \sqrt 5 } }}\\
= \dfrac{{\left( {6 + 2\sqrt 5 } \right)\sqrt {6 - 2\sqrt 5 } }}{{2.\sqrt {6 + 2\sqrt 5 } }}\\
= \dfrac{{{{\left( {\sqrt 5 + 1} \right)}^2}\sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} }}{{2.\sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} }}\\
= \dfrac{{{{\left( {\sqrt 5 + 1} \right)}^2}.\left( {\sqrt 5 - 1} \right)}}{{2.\left( {\sqrt 5 + 1} \right)}}\\
= \dfrac{{\left( {\sqrt 5 + 1} \right).\left( {\sqrt 5 - 1} \right)}}{2}\\
= \dfrac{{5 - 1}}{2}\\
= 2\\
B4)Dkxd:\left\{ \begin{array}{l}
{x^2} - 9 \ge 0\\
x - 3 \ge 0
\end{array} \right. \Leftrightarrow x \ge 3\\
\sqrt {{x^2} - 9} - \sqrt {x - 3} = 0\\
\Leftrightarrow \sqrt {x - 3} .\sqrt {x + 3} - \sqrt {x - 3} = 0\\
\Leftrightarrow \sqrt {x - 3} \left( {\sqrt {x + 3} - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {x - 3} = 0\\
\sqrt {x + 3} - 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x + 3 = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = - 2\left( {ktm} \right)
\end{array} \right.\\
Vay\,x = 3
\end{array}$
Bài 5 xem lại đề, tại sao là căn x-x?
