Đáp án:
$\begin{array}{l}
3)\\
\dfrac{1}{{\sqrt 1 + \sqrt 2 }} + \dfrac{1}{{\sqrt 2 + \sqrt 3 }} + \dfrac{1}{{\sqrt 3 + \sqrt 4 }} + ... + \dfrac{1}{{\sqrt {n - 1} + \sqrt n }}\\
= \dfrac{{\sqrt 2 - \sqrt 1 }}{{2 - 1}} + \dfrac{{\sqrt 3 - \sqrt 2 }}{{3 - 2}} + \dfrac{{\sqrt 4 - \sqrt 3 }}{{4 - 3}} + .. + \dfrac{{\sqrt n - \sqrt {n - 1} }}{{n - n + 1}}\\
= \sqrt 2 - 1 + \sqrt 3 - \sqrt 2 + \sqrt 4 - \sqrt 3 + .. + \sqrt n - \sqrt {n - 1} \\
= \sqrt n - 1\\
4)Dkxd:\left\{ \begin{array}{l}
x \ge - 1\\
y \ge 3\\
z \ge 1
\end{array} \right.\\
\sqrt {x + 1} + \sqrt {y - 3} + \sqrt {z - 1} = \dfrac{1}{2}\left( {x + y + z} \right)\\
\Leftrightarrow x + y + z = 2\sqrt {x + 1} + 2\sqrt {y - 3} + 2\sqrt {z - 1} \\
\Leftrightarrow x - 2\sqrt {x + 1} + y - 2\sqrt {y - 3} + z - 2\sqrt {z - 1} = 0\\
\Leftrightarrow \left( {x + 1 - 2\sqrt {x + 1} + 1} \right) + \left( {y - 3 - 2\sqrt {y - 3} + 1} \right)\\
+ \left( {z - 1 - 2\sqrt {z - 1} + 1} \right) = 0\\
\Leftrightarrow {\left( {\sqrt {x + 1} - 1} \right)^2} + {\left( {\sqrt {y - 3} - 1} \right)^2} + {\left( {\sqrt {z - 1} - 1} \right)^2} = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
\left( {\sqrt {x + 1} - 1} \right) = 0\\
\sqrt {y - 3} - 1 = 0\\
\sqrt {z - 1} - 1 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\sqrt {x + 1} = 1\\
\sqrt {y - 3} = 1\\
\sqrt {z - 1} = 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x + 1 = 1 \Leftrightarrow x = 0\left( {tm} \right)\\
y - 3 = 1 \Leftrightarrow y = 4\left( {tm} \right)\\
z - 1 = 1 \Leftrightarrow z = 2\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = 0;y = 4;z = 2
\end{array}$
