$a)$ $\sqrt[]{4x²}$ $=8$

$⇔2║x║=8$

$⇔║x║=4$

$⇔x=4$ $hoặc$ $x=-4$
$b)$ $\sqrt[]{16x²}$ $=║-20║$

⇔ $4║x║=20$

⇔$║x║=5$
⇔$ x=5$ $hoặc$ $x=-5$

$c)$ $\sqrt[]{x²+4x+4}$ $=2$

⇔ $\sqrt[]{(x+2)²}$ $=2$

⇔ $║x+2║=2$

$⇔ x+2=2$ $hoặc$ $x+2=-2$

$⇔ x=0$ $hoặc$ $x=-4$

$d)$ $\sqrt[]{25x²-10x+1}$ $=4x-9$

⇔ $\sqrt[]{(5x-1)²}$ $=4x-9$

⇔ $║5x-1║=4x-9$

$⇔ 5x-1=4x-9$ $hoặc$ $1-5x=4x-9$

$⇔ x=-8$ $hoặc$ $x=9/10$

$e)$ $\sqrt[]{x²-6x+9}$ $=4-x$

⇔ $\sqrt[]{(x-3)²}$ $=4-x$

⇔$║x-3║=4-x$

⇔$x-3=4-x$ $hoặc$ $3-x=4-x$

⇔$x=7/2$ $hoặc$ $3=4(vô$ $lý)$

$g)$ $\sqrt[]{x+4\sqrt[]{x-4}}$ $=2$

⇔ $\sqrt[]{x-4+2.2\sqrt[]{x-4}+2²}$ $=2$

⇔ $\sqrt[]{(\sqrt[]{x-4}+2)²}$ $=2$

⇔ ║$\sqrt[]{x-4}$+$2║=2$

⇔ $\sqrt[]{x-4}$+$2=2$

⇔ $\sqrt[]{x-4}$ $=0$

⇔ $x-4=0$

⇔ $x=4$

Đáp án:

  a,  $\sqrt[2]{4x²}$ = 8

⇒ $\sqrt[2]{(2²)(±x)²}$ = 8

⇒ 2. ±x = 8

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

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

 Vậy x = 4 hoặc x = -4

  b, $\sqrt[2]{16x²}$ = | -20 |

⇒ $\sqrt[2]{4².x²}$ = 20  ( do biểu thức nằm trong dấu g/trị tuyệt đối luôn ≥ 0 )

⇒ ±x . 4 = 20

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

  Vậy x = 5 hoặc x = -5

 c, $\sqrt[2]{x² + 4x + 2}$ = 2

⇒ $\sqrt[2]{(x + 2 )²}$ = 2

⇒ | x + 2 | = 2

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

⇒ \(\left[ \begin{array}{l}x=0\\x=-4\end{array} \right.\) 

  Vậy x = 0 hoặc x = -4

d, $\sqrt[2]{25x² - 10x + 1}$ = 4x - 9

⇒ $\sqrt[2]{( 5x - 1 )²}$ = 4x - 9

⇒ | 5x - 1 | = 4x - 9

⇒ \(\left[ \begin{array}{l}5x - 1 = 4x - 9\\5x - 1 =-4x + 9\end{array} \right.\) 

⇒ \(\left[ \begin{array}{l}x=-8\\x=-$\frac{8}{9}$ \end{array} \right.\) 

e,  $\sqrt[2]{x² - 6x + 9}$ = 4 - x

⇒ $\sqrt[2]{( x - 3 )²}$ = 4 - x

⇒ | x - 3 | = 4 - x

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

⇒ \(\left[ \begin{array}{l}x=$\frac{7}{2}$ \\0 = -1\end{array} \right.\) ( loại )

 Vậy x = $\frac{7}{2}$ 

f, $\sqrt[2]{x + 2$\sqrt[2]{x - 1}$ }$ = 4

⇒ x + 2$\sqrt[2]{x - 1}$ = 4

⇒ 2$\sqrt[2]{x - 1} = 4 - x

⇒ 4 . ( x - 1 ) = ( 4 - x )²

⇒ 4x - 4 = 16 - 8x + x²

⇒ x² - 12x + 20 = 0

⇒ ( x - 10 ) ( x - 2 ) = 0

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

 Vậy x = 10 hoặc x = -2

g , $\sqrt[2]{x + 4$\sqrt[2]{x - 4}$ }$ = 2

⇒ $\sqrt[2]{x - 4 + 2.2$\sqrt[2]{x - 4}$ + 2²}$ = 2

⇒ $\sqrt[2]{( $\sqrt[2]{x - 4}$ + 2)²}$ = 2 

⇒ | $\sqrt[2]{x - 4}$ + 2 | = 2

⇒ \(\left[ \begin{array}{l}$\sqrt[2]{x - 4}$ + 2=2\\$\sqrt[2]{x - 4}$ + 2= -2\end{array} \right.\) 

⇒ \(\left[ \begin{array}{l}$\sqrt[2]{x - 4}$ = 0\\$\sqrt[2]{x - 4}$ = -4\end{array} \right.\) 

⇒ \(\left[ \begin{array}{l}x - 4=0\\x - 4=-2\end{array} \right.\) 

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