Đáp án:
$\begin{array}{l}
d)\\
x = \dfrac{1}{{1 + \sqrt 3 }} + \dfrac{1}{{\sqrt 3 + \sqrt 5 }} + ... + \dfrac{1}{{\sqrt {79} + \sqrt {81} }}\\
= \dfrac{{\sqrt 3 - 1}}{{3 - 1}} + \dfrac{{\sqrt 5 - \sqrt 3 }}{{5 - 3}} + ... + \dfrac{{\sqrt {81} - \sqrt {79} }}{{81 - 79}}\\
= \dfrac{{\sqrt 3 - 1 + \sqrt 5 - \sqrt 3 + ... + \sqrt {81} - \sqrt {79} }}{2}\\
= \dfrac{{\sqrt {81} - 1}}{2}\\
= \dfrac{{9 - 1}}{2}\\
= 4\\
e)Dkxd:x \ge 1\\
\sqrt {2{x^2} - 3x - 5} = x - 1\\
\Leftrightarrow 2{x^2} - 3x - 5 = {x^2} - 2x + 1\\
\Leftrightarrow {x^2} - x - 6 = 0\\
\Leftrightarrow \left( {x - 3} \right)\left( {x + 2} \right) = 0\\
\Leftrightarrow x = 3\left( {tmdk} \right)\\
Vay\,x = 3\\
f)Dkxd:x \ge - \dfrac{1}{3}\\
\left| {2x - 6} \right| = 3x + 1\\
\Leftrightarrow \left[ \begin{array}{l}
2x - 6 = 3x + 1\\
2x - 6 = - 3x - 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 7\left( {ktm} \right)\\
5x = 5
\end{array} \right.\\
\Leftrightarrow x = 1\left( {tmdk} \right)\\
Vậy\,x = 1
\end{array}$
