Đáp án:

$a. \sqrt[]{x^{2}-2x+1} = 2x - 4$ $( x ≥ 2 )$

⇔ $\sqrt[]{(x-1)^{2}} = 2x - 4$

⇔ $| x - 1 | = 2x-4$

⇔ \(\left[ \begin{array}{l}x-1=2x-4\\x-1=-2x+4\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=3\\x=\frac{5}{3}\end{array} \right.\) 

Ma $x ≥ 2 ⇒ x = 3$

$b. \sqrt[]{x^{2}+6x+9} = 5 - x$ $( x ≤ 5 )$

⇔ $\sqrt[]{(x+3)^{2}} = 5 - x$

⇔ $| x + 3 | = 5 - x$

⇔ \(\left[ \begin{array}{l}x+3=5-x\\x+3=-5+x\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}2x=2\\0x=-8\end{array} \right.\) 

⇔ $x = 1$ ( TM )

$c. \sqrt[]{4x^{2}-4x+1} - 5 = 3x$

⇔ $\sqrt[]{(2x-1)^{2}} = 3x + 5$ $( x ≥ - \frac{5}{3} )$

⇔ $| 2x - 1 | = 3x + 5$

⇔ \(\left[ \begin{array}{l}2x-1=3x+5\\2x-1=-3x-5\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=-6\\5x=-4\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=-6\\x=-\frac{4}{5}\end{array} \right.\) 

Ma $x ≥ - \frac{5}{3} ⇒ x = - \frac{4}{5}$

$d. \sqrt[]{9x^{2}-12x+4} + 4x = 12$

⇔ $\sqrt[]{(3x-2)^{2}} = 12 - 4x$ $( x ≤ 3 )$

⇔ $| 3x - 2 | = 12 - 4x$

⇔ \(\left[ \begin{array}{l}3x-2=12-4x\\3x-2=-12+4x\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}7x=14\\x=10\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=2\\x=10\end{array} \right.\) 

Ma $x ≤ 3 ⇒ x = 2$

$e. \sqrt[]{25x^{2}-20x+4} - 3x = 9$

⇔ $\sqrt[]{(5x-2)^{2}} = 9 + 3x$ $( x ≥ - 3 )$

⇔ $| 5x - 2 | = 9 + 3x$

⇔ \(\left[ \begin{array}{l}5x-2=9+3x\\5x-2=-9-3x\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}2x=11\\8x=-7\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=\frac{11}{2}\\x=-\frac{7}{8}\end{array} \right.\) ( TM )

$f. \sqrt[]{16x^{2}+24x+9} + 7 = 21x$

⇔ $\sqrt[]{(4x+3)^{2}} = 21x - 7$ $( x ≥ \frac{1}{3} )$

⇔ $| 4x + 3 | = 21x - 7$

⇔ \(\left[ \begin{array}{l}4x+3=21x-7\\4x+3=-21x+7\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}17x=10\\25x=4\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=\frac{10}{17}\\x=\frac{4}{25}\end{array} \right.\) 

Ma $x ≥ \frac{1}{3} ⇒ x = \frac{10}{17}$