Đáp án:
$\begin{array}{l}
B3)\\
a)A = \sqrt {17 - 3\sqrt {32} } \\
= \sqrt {9 - 2.3.2\sqrt 2 + 8} \\
= \sqrt {{{\left( {3 - 2\sqrt 2 } \right)}^2}} \\
= 3 - 2\sqrt 2 \\
b)B = \sqrt {5 + 2\sqrt 6 } - \sqrt {5 - 2\sqrt 6 } \\
= \sqrt {{{\left( {\sqrt 3 + \sqrt 2 } \right)}^2}} - \sqrt {{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}} \\
= \sqrt 3 + \sqrt 2 - \left( {\sqrt 3 - \sqrt 2 } \right)\\
= 2\sqrt 2 \\
B4)a)\\
A = \sqrt {{x^2} - 2x + 1} + \sqrt {{x^2} + 4x + 4} \\
= \sqrt {{{\left( {x - 1} \right)}^2}} + \sqrt {{{\left( {x + 2} \right)}^2}} \\
= \left| {x - 1} \right| + \left| {x + 2} \right|\\
+ Khi:x \ge 1\\
\Leftrightarrow A = x - 1 + x + 2 = 2x + 1\\
+ Khi: - 2 \le x < 1\\
\Leftrightarrow A = 1 - x + x + 2 = 3\\
+ Khi:x < - 2\\
\Leftrightarrow A = 1 - x - x - 2 = - 2x - 1\\
b)A = 7\\
\Leftrightarrow \left[ \begin{array}{l}
2x + 1 = 7\left( {khi:x \ge 1} \right)\\
- 2x - 1 = 7\left( {khi:x < - 2} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 3\left( {tm} \right)\\
x = - 4\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = 3;x = - 4\\
B5)\\
B = \sqrt {x + 2\sqrt {x - 1} } + \sqrt {x + 3 + 2\sqrt {x + 2} } \\
= \sqrt {x - 1 + 2\sqrt {x - 1} + 1} + \sqrt {x + 2 + 2\sqrt {x + 2} + 1} \\
= \sqrt {{{\left( {\sqrt {x - 1} + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt {x + 2} + 2} \right)}^2}} \\
= \sqrt {x - 1} + 1 + \sqrt {x + 2} + 2\\
= \sqrt {x - 1} + \sqrt {x + 2} + 3
\end{array}$
