Đáp án:
\(\begin{array}{l}
a)6\sqrt 5 + 25\\
b)x \in \emptyset
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\dfrac{{2 + 3\sqrt 5 }}{{\sqrt 5 - 2}} + \dfrac{4}{{\sqrt 5 + 3}} + \dfrac{{\sqrt 5 \left( {3 - \sqrt 5 } \right)}}{{\sqrt 5 }}\\
= \dfrac{{\left( {2 + 3\sqrt 5 } \right)\left( {\sqrt 5 + 2} \right)}}{{5 - 4}} + \dfrac{{4\left( {\sqrt 5 - 3} \right)}}{{5 - 9}} + 3 - \sqrt 5 \\
= 2\sqrt 5 + 4 + 15 + 6\sqrt 5 - \sqrt 5 + 3 + 3 - \sqrt 5 \\
= 6\sqrt 5 + 25\\
b)DK:x \ge - \dfrac{3}{4}\\
\dfrac{{\sqrt {4x + 3} }}{{\sqrt {x + 1} }} = 3\\
\to \sqrt {4x + 3} = 3\sqrt {x + 1} \\
\to 4x + 3 = 9\left( {x + 1} \right)\\
\to 5x = - 6\\
\to x = - \dfrac{6}{5}\left( {KTM} \right)\\
\to x \in \emptyset
\end{array}\)
