1) √4x-1 – √4x-2√3 = 0
2) -√9+4√5 + √5x+2 = 0
3) √50x-25 + √4x+4 = √16x+16 – √32x-16
4) √x^2+2x√5+5 – √6-2√5 = 0
1) √4x-1 – √4x-2√3 = 0
2) -√9+4√5 + √5x+2 = 0
3) √50x-25 + √4x+4 = √16x+16 – √32x-16
4) √x^2+2x√5+5 – √6-2√5 = 0
\[\begin{array}{l}
1)\,\,\sqrt {4x – 1} – \sqrt {4x} – 2\sqrt 3 = 0\,\,\,xem\,\,\,lai\,\,\,de\,\,\,bai.\\
2)\,\, – \sqrt {9 + 4\sqrt 5 } + \sqrt {5x + 2} = 0\\
DK:\,\,\,x \ge – \frac{2}{5}\\
pt \Leftrightarrow \sqrt {5x + 2} = \sqrt {9 + 4\sqrt 5 } \\
\Leftrightarrow 5x + 2 = 9 + 4\sqrt 5 \\
\Leftrightarrow 5x = 7 + 4\sqrt 5 \\
\Leftrightarrow x = \frac{{7 + 4\sqrt 5 }}{2}\,\,\,\left( {tm} \right).\\
3)\,\,\,\sqrt {50x – 25} + \sqrt {4x + 4} = \sqrt {16x + 16} – \sqrt {32x – 16} \\
DK:\,\,\,x \ge \frac{1}{2}\\
pt \Leftrightarrow \sqrt {25\left( {2x – 1} \right)} + \sqrt {4\left( {x + 1} \right)} = \sqrt {16\left( {x + 1} \right)} – \sqrt {16\left( {2x – 1} \right)} \\
\Leftrightarrow 5\sqrt {2x – 1} + 4\sqrt {2x – 1} = 4\sqrt {x + 1} – 2\sqrt {x + 1} \\
\Leftrightarrow 9\sqrt {2x – 1} = 2\sqrt {x + 1} \\
\Leftrightarrow 81\left( {2x – 1} \right) = 4\left( {x + 1} \right)\\
\Leftrightarrow 162x – 81 = 4x + 4\\
\Leftrightarrow 158x = 89\\
\Leftrightarrow x = \frac{{89}}{{158}}\,\,\,\left( {tm} \right)\\
4)\,\,\,\sqrt {{x^2} + 2x\sqrt 5 + 5} – \sqrt {6 – 2\sqrt 5 } = 0\\
\Leftrightarrow \sqrt {{{\left( {x + \sqrt 5 } \right)}^2}} = \sqrt {6 – 2\sqrt 5 } \\
\Leftrightarrow \sqrt {{{\left( {x + \sqrt 5 } \right)}^2}} = \sqrt {{{\left( {\sqrt 5 – 1} \right)}^2}} \\
\Leftrightarrow \left[ \begin{array}{l}
x + \sqrt 5 = \sqrt 5 – 1\\
x + \sqrt 5 = 1 – \sqrt 5
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = – 1\\
x = 1 – 2\sqrt 5
\end{array} \right..
\end{array}\]