Đáp án:
$\begin{array}{l}
1)a)Dkxd:\left( {{x^2} - 5x - 6} \right) > 0\\
\Rightarrow \left( {x - 6} \right)\left( {x + 1} \right) > 0\\
\Rightarrow \left[ \begin{array}{l}
x > 6\\
x < - 1
\end{array} \right.\\
Vay\,x > 6\,hoac\,x < - 1\\
b)Dkxd:{x^2} - 4 > 0\\
\Rightarrow \left( {x - 2} \right)\left( {x + 2} \right) > 0\\
\Rightarrow \left[ \begin{array}{l}
x > 2\\
x < - 2
\end{array} \right.\\
Vay\,x > 2\,hoac\,x < - 2\\
2)a)y = \sqrt[3]{{2x - 1}}\\
\Rightarrow y' = \dfrac{1}{3}.2.{\left( {2x - 1} \right)^{\dfrac{{ - 2}}{3}}}\\
= \dfrac{2}{{3.\sqrt[3]{{{{\left( {2x - 1} \right)}^2}}}}}\\
b)y = {\left( {3x + 1} \right)^{ - 5}}\\
\Rightarrow y' = - 5.3.{\left( {3x + 1} \right)^{ - 6}}\\
\Rightarrow y' = \dfrac{{ - 15}}{{{{\left( {3x + 1} \right)}^6}}}\\
c)y = \dfrac{{\sqrt[5]{{x + 1}}}}{{\sqrt {x - 1} }}\\
\Rightarrow y' = \dfrac{{\dfrac{1}{5}.{{\left( {x + 1} \right)}^{\dfrac{{ - 4}}{5}}}.\sqrt {x - 1} - \dfrac{1}{{2\sqrt {x - 1} }}.\sqrt[5]{{x + 1}}}}{{x - 1}}\\
= \dfrac{{\dfrac{{\sqrt {x - 1} }}{{5\sqrt[5]{{{{\left( {x + 1} \right)}^4}}}}} - \dfrac{{\sqrt[5]{{x + 1}}}}{{2\sqrt {x - 1} }}}}{{x - 1}}
\end{array}$
