Em tách từng câu ra hỏi thôi nhé
\(\begin{array}{l}
1)\,A = \frac{{3 + \sqrt 5 - \left( {3 - \sqrt 5 } \right)}}{{\left( {3 + \sqrt 5 } \right)\left( {3 - \sqrt 5 } \right)}}:\sqrt 5 \\
= \frac{{2\sqrt 5 }}{{9 - 5}}.\frac{1}{{\sqrt 5 }} = \frac{1}{2}\\
b)\,B = \sqrt {16.3} + \sqrt {\frac{{16}}{3}} + 2\sqrt {25.3} - 5\sqrt {\frac{4}{3}} \\
= 4\sqrt 3 + \frac{{4\sqrt 3 }}{3} + 10\sqrt 3 - \frac{{10\sqrt 3 }}{3}\\
= 14\sqrt 3 - 2\sqrt 3 = 12\sqrt 3 \\
2)\,a)\,DK:\,x \le 1\\
pt \Leftrightarrow \sqrt {1 - x} + 2\sqrt {1 - x} - 12 = 0\\
\Leftrightarrow 3\sqrt {1 - x} = 12\\
\Leftrightarrow \sqrt {1 - x} = 4\\
\Leftrightarrow 1 - x = 16\\
\Leftrightarrow x = - 15\left( {tm} \right)\\
b)\,\sqrt {4{x^2} - 4x + 1} = 3\\
\Leftrightarrow \sqrt {{{\left( {2x - 1} \right)}^2}} = 3\\
\Leftrightarrow {\left( {2x - 1} \right)^2} = 9\\
\Leftrightarrow \left[ \begin{array}{l}
2x - 1 = 3\\
2x - 1 = - 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = - 1
\end{array} \right.
\end{array}\)
