Giải thích các bước giải:
$\eqalign{ & a)\sqrt {3{x^2} - 2x - 1} - 3x = 1 \cr & ĐKXĐ:\,3{x^2} - 2x - 1 \geqslant 0 \cr & pt \Leftrightarrow \sqrt {3{x^2} - 2x - 1} = 3x + 1 \cr & \Leftrightarrow 3x + 1 \geqslant 0;\,3{x^2} - 2x - 1 = 9{x^2} + 6x + 1 \cr & \Leftrightarrow x \geqslant - \frac{1}{3};\,6{x^2} + 8x + 2 = 0 \cr & \Leftrightarrow x \geqslant - \frac{1}{3};\,x = - \frac{1}{3}\,hoặc\,x = - 1 \cr & \Leftrightarrow x = - \frac{1}{3}\,hoặc\,x = - 1(tm\,ĐKXĐ) \cr} $
$\eqalign{ & b)\sqrt {3x - 2} + \sqrt {x - 1} = \sqrt {5x - 1} \cr & DKXD:x\, \geqslant 1 \cr & pt \Leftrightarrow {(\sqrt {3x - 2} + \sqrt {x - 1} )^2} = 5x - 1 \cr & \Leftrightarrow 3x - 2 + x - 1 + 2\sqrt {(x - 1)(3x - 2)} = 5x - 1 \cr & \Leftrightarrow 2\sqrt {(x - 1)(3x - 2)} = x + 2 \cr & \Leftrightarrow 4(x - 1)(3x - 2) = {x^2} + 4x + 4(do\,x\, \geqslant 1\,nên\,x + 2 > 0) \cr & \Leftrightarrow 12{x^2} - 20x + 8 = \,{x^2} + 4x + 4 \cr & \Leftrightarrow 11{x^2} - 24x + 4 = 0 \cr & \Leftrightarrow 11(x - 2)(x - \frac{2}{{11}}) = 0 \cr & \Leftrightarrow x = 2\,hoặc\,x = \frac{2}{{11}} \cr} $
