Bài 2 (2 điểm). Giải các phương trình sau: a) $\sqrt{4x+12} - 5\sqrt{\frac{9x+27}{100}}+ \sqrt{x+3} -6 = 0$ b)  $\sqrt{x^2-4x+4}-1 =2x$ c)  $3\sqrt{x+2} -\sqrt{x^2-4} =0$

Giải thích các bước giải:

a) ĐKXĐ: `x≥-3`

$\begin{array}{l} \sqrt {4x + 12} - 5\sqrt {\dfrac{{9x + 27}}{{100}}} + \sqrt {x + 3} - 6 = 0\\ \Leftrightarrow \sqrt {4\left( {x + 3} \right)} - 5\dfrac{{\sqrt {9\left( {x + 3} \right)} }}{{10}} + \sqrt {x + 3} - 6 = 0\\ \Leftrightarrow 2\sqrt {x + 3} - \dfrac{3}{2}\sqrt {x + 3} + \sqrt {x + 3} - 6 = 0\\ \Leftrightarrow \dfrac{3}{2}\sqrt {x + 3} = 6\\ \Leftrightarrow \sqrt {x + 3} = 4\\ \Leftrightarrow x + 3 = 16\\ \Leftrightarrow x = 13(TM) \end{array}$

Vậy `x=13`.

b) ĐKXĐ: `x≥ -1/2`

$\begin{array}{l} \sqrt {{x^2} - 4x + 4} - 1 = 2x\\ \Leftrightarrow \sqrt {{{\left( {x - 2} \right)}^2}} = 2x + 1\\ \Leftrightarrow \left| {x - 2} \right| = 2x + 1\\ \Leftrightarrow \left[ \begin{array}{l} x - 2 = 2x + 1\\ 2 - x = 2x + 1 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = - 3(KTM)\\ x = \dfrac{1}{3}(TM) \end{array} \right. \end{array}$

Vậy `x=1/3`.

c) ĐKXĐ: $\left\{ \begin{array}{l} x + 2 \ge 0\\ {x^2} - 4 \ge 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x + 2 \ge 0\\ \left( {x - 2} \right)\left( {x + 2} \right) \ge 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x + 2 \ge 0\\ x - 2 \ge 0 \end{array} \right. \Leftrightarrow x \ge 2$

$\begin{array}{l} 3\sqrt {x + 2} - \sqrt {{x^2} - 4} = 0\\ \Leftrightarrow 3\sqrt {x + 2} - \sqrt {\left( {x - 2} \right)\left( {x + 2} \right)} = 0\\ \Leftrightarrow \sqrt {x + 2} \left( {3 - \sqrt {x - 2} } \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sqrt {x + 2} = 0\\ 3 - \sqrt {x - 2} = 0 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x + 2 = 0\\ \sqrt {x - 2} = 3 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = - 2(KTM)\\ x - 2 = 6 \end{array} \right.\\ \Leftrightarrow x = 8(TM) \end{array}$

Vậy `x=8`.