Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge - 1\\
\sqrt {x + 1} = 3\\
\Leftrightarrow x + 1 = 9\\
\Leftrightarrow x = 8\left( {tmdk} \right)\\
Vậy\,x = 8\\
b)Dkxd:{x^2} - 3 \ge 0 \Leftrightarrow {x^2} \ge 3\\
\sqrt {{x^2} - 3} - 1 = \sqrt 9 \\
\Leftrightarrow \sqrt {{x^2} - 3} = 3 + 1 = 4\\
\Leftrightarrow {x^2} - 3 = 16\\
\Leftrightarrow {x^2} = 19\left( {tmdk} \right)\\
\Leftrightarrow x = \pm \sqrt {19} \\
Vậy\,x = \pm \sqrt {19} \\
c)\sqrt {{{\left( {x - 3} \right)}^4}} = 16\\
\Leftrightarrow {\left( {x - 3} \right)^2} = 4\\
\Leftrightarrow \left[ \begin{array}{l}
x - 3 = 2\\
x - 3 = - 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 5\\
x = 1
\end{array} \right.\\
Vậy\,x = 1;x = 5\\
d)\sqrt {{{\left( {2x + 3} \right)}^2}} = 7 - \sqrt 9 \\
\Leftrightarrow \left| {2x + 3} \right| = 7 - 3 = 4\\
\Leftrightarrow \left[ \begin{array}{l}
2x + 3 = 4\\
2x + 3 = - 4
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = 1\\
2x = - 7
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = - \dfrac{7}{2}
\end{array} \right.\\
Vậy\,x = \dfrac{1}{2};x = - \dfrac{7}{2}\\
B4)\\
a)\sqrt {27} < \sqrt {31} \\
b)7 = \sqrt {49} > \sqrt {48} \\
c)2\sqrt {99} > 2.\sqrt {25} = 2.5 = 10\\
d)\sqrt 3 + \sqrt {15} \\
< \sqrt 5 + \sqrt {16} \\
\Leftrightarrow \sqrt 3 + \sqrt {15} < \sqrt 5 + 4\\
e)\sqrt 7 + \sqrt {15} < \sqrt 9 + \sqrt {16} = 3 + 4 = 7
\end{array}$
