Đáp án:
$\begin{array}{l}
B3)a)Dkxd:x \ge 0\\
2\sqrt {3x} - 4\sqrt {3x} = 27 - 3\sqrt {3x} \\
\Leftrightarrow 2\sqrt {3x} - 4\sqrt {3x} + 3\sqrt {3x} = 27\\
\Leftrightarrow \sqrt {3x} = 27\\
\Leftrightarrow 3x = {27^2}\\
\Leftrightarrow x = 243\\
Vậy\,x = 243\\
b)Dkxd:x \le 3\\
\sqrt {{x^2} + 2x + 1} = 3 - x\\
\Leftrightarrow \sqrt {{{\left( {x + 1} \right)}^2}} = 3 - x\\
\Leftrightarrow \left[ \begin{array}{l}
x + 1 = 3 - x\\
x + 1 = x - 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\left( {tm} \right)\\
1 = - 3\left( {ktm} \right)
\end{array} \right.\\
Vậy\,x = 1\\
B4)\\
a)2 + \sqrt 6 = \sqrt 4 + \sqrt 6 > \sqrt 3 + \sqrt 5 \\
b)\sqrt {1234} + \sqrt {1236} - 2\sqrt {1235} \\
= \sqrt {1236} - \sqrt {1235} - \left( {\sqrt {1235} - \sqrt {1234} } \right)\\
= \dfrac{1}{{\sqrt {1236} + \sqrt {1235} }} - \dfrac{1}{{\sqrt {1235} + \sqrt {1234} }} < 0\\
\Leftrightarrow \sqrt {1234} + \sqrt {1236} < 2\sqrt {1235} \\
B5)\\
a)x \ge 0;x \ne 4\\
T = \dfrac{2}{{\sqrt x + 3}} + \dfrac{5}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 2} \right)}} - \dfrac{1}{{\sqrt x - 2}}\\
= \dfrac{{2\left( {\sqrt x - 2} \right) + 5 - \left( {\sqrt x + 3} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{2\sqrt x - 4 + 5 - \sqrt x - 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\sqrt x - 2}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{1}{{\sqrt x + 3}}\\
b)T = \dfrac{1}{4}\\
\Leftrightarrow \dfrac{1}{{\sqrt x + 3}} = \dfrac{1}{4}\\
\Leftrightarrow \sqrt x + 3 = 4\\
\Leftrightarrow \sqrt x = 1\\
\Leftrightarrow x = 1\left( {tmdk} \right)\\
Vậy\,x = 1
\end{array}$
