Giải thích các bước giải:
$\begin{array}{l}
a)\dfrac{{\sqrt {15} + 1}}{{\sqrt 3 }}\left( {\dfrac{{ - 4}}{{\sqrt 3 + \sqrt 5 }} + \dfrac{{ - 4}}{{\sqrt 5 + \sqrt 7 }} + ... + \dfrac{{ - 4}}{{\sqrt {43} + \sqrt {45} }}} \right)\\
= \dfrac{{ - 2\left( {\sqrt {15} + 1} \right)}}{{\sqrt 3 }}\left( {\dfrac{2}{{\sqrt 3 + \sqrt 5 }} + \dfrac{2}{{\sqrt 5 + \sqrt 7 }} + ... + \dfrac{2}{{\sqrt {43} + \sqrt {45} }}} \right)\\
= \dfrac{{ - 2\left( {\sqrt {15} + 1} \right)}}{{\sqrt 3 }}\left( {\dfrac{{5 - 3}}{{\sqrt 3 + \sqrt 5 }} + \dfrac{{7 - 5}}{{\sqrt 5 + \sqrt 7 }} + ... + \dfrac{{45 - 43}}{{\sqrt {43} + \sqrt {45} }}} \right)\\
= \dfrac{{ - 2\left( {\sqrt {15} + 1} \right)}}{{\sqrt 3 }}\left( {\sqrt 5 - \sqrt 3 + \sqrt 7 - \sqrt 5 + ... + \sqrt {45} - \sqrt {43} } \right)\\
= \dfrac{{ - 2\left( {\sqrt {15} + 1} \right)}}{{\sqrt 3 }}\left( {\sqrt {45} - \sqrt 3 } \right)\\
= \dfrac{{ - 2\left( {\sqrt {15} + 1} \right)}}{{\sqrt 3 }}.\sqrt 3 \left( {\sqrt {15} - 1} \right)\\
= - 2\left( {\sqrt {15} + 1} \right)\left( {\sqrt {15} - 1} \right)\\
= - 2\left( {15 - 1} \right)\\
= - 28
\end{array}$
$\begin{array}{l}
b)\sqrt {{x^2} - 3x + 2} + \sqrt {x + 3} = \sqrt {{x^2} + 2x - 3} + \sqrt {x - 2} \left( {DK:x \ge 2} \right)\\
\Leftrightarrow \sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} + \sqrt {x + 3} = \sqrt {\left( {x - 1} \right)\left( {x + 3} \right)} + \sqrt {x - 2} \\
\Leftrightarrow \sqrt {x - 1} .\sqrt {x - 2} - \sqrt {x - 1} .\sqrt {x + 3} + \sqrt {x + 3} - \sqrt {x - 2} = 0\\
\Leftrightarrow \left( {\sqrt {x + 3} - \sqrt {x - 2} } \right)\left( {1 - \sqrt {x - 1} } \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {x + 3} - \sqrt {x - 2} = 0\\
1 - \sqrt {x - 1} = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {x + 3} = \sqrt {x - 2} \\
\sqrt {x - 1} = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x + 3 = x - 2\\
x - 1 = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
3 = - 2\left( {mt} \right)\\
x = 2\left( c \right)
\end{array} \right.\\
\Leftrightarrow x = 2
\end{array}$
Vậy phương trình có tập nghiệm là $S = \left\{ 2 \right\}$
