Đáp án:

$x = \{1;4;16;25;49\}$

Giải thích các bước giải:

$\begin{array}{l}1)\quad \sqrt{x^2 +4x + 4} = 6\\ \to \sqrt{(x+2)^2} = 6\\ \to |x+2| = 6\\ \to \left[\begin{array}{l}x+2 = 6\\x + 2 =-6\end{array}\right.\\ \to \left[\begin{array}{l}x =-4\\x = -8\end{array}\right.\\ 2)\quad A = \dfrac{1}{\sqrt x + 3} + \dfrac{3}{\sqrt x - 3} - \dfrac{6}{9-x}\qquad (x \geq 0;\, x \ne9)\\ \to A = \dfrac{\sqrt x - 3}{(\sqrt x - 3)(\sqrt x+3)} + \dfrac{3(\sqrt x+3)}{(\sqrt x - 3)(\sqrt x+3)} + \dfrac{6}{(\sqrt x - 3)(\sqrt x+3)}\\ \to A = \dfrac{\sqrt x - 3 + 3\sqrt x + 9 + 6}{(\sqrt x - 3)(\sqrt x+3)}\\ \to A = \dfrac{4\sqrt x +12}{(\sqrt x - 3)(\sqrt x+3)}\\ \to A = \dfrac{4(\sqrt x +3)}{(\sqrt x - 3)(\sqrt x+3)}\\ \to A =\dfrac{4}{\sqrt x-3}\\ Với\,\,x=4\,\,ta\,\,được:\\ A = \dfrac{4}{\sqrt4 -3} = -4\\ A \in \Bbb Z\Leftrightarrow \dfrac{4}{\sqrt x - 3}\in \Bbb Z\\ \sqrt x - 3 \in Ư(4)=\{-4;-2;-1;1;2;4\}\\ Do\,\,\sqrt x \geq 0 \to \sqrt x - 3 \geq -3\\ nên\,\,\sqrt x - 3 =\{-2;-1;1;2;4\}\\ \to \sqrt x =\{1;2;4;5;7\}\\ \to x = \{1;4;16;25;49\} \end{array}$

Đáp án:

 1. $\sqrt{x^2 + 4x + 4} = 6$ 

$\sqrt{(x + 2)^2} = 6 \to \left | x + 2  \right | = 6$ 

Suy ra: 

*) $x + 2 = 6 \to x = 4$ 

*) $x + 2 = - 6 \to x = - 8$
2. ĐKXĐ: $x \geq 0$;    $x \neq 9$

a. $A = \dfrac{1}{\sqrt{x} + 3} + \dfrac{3}{\sqrt{x} - 3} + \dfrac{6}{(\sqrt{x} + 3)(\sqrt{x} - 3)}$ 

$A = \dfrac{\sqrt{x} - 3 + 3(\sqrt{x} + 3) + 6}{(\sqrt{x} + 3)(\sqrt{x} - 3)}$ 

$A = \dfrac{4\sqrt{x} + 12}{(\sqrt{x} + 3)(\sqrt{x} - 3)} = \dfrac{4(\sqrt{x} + 3)}{(\sqrt{x} + 3)(\sqrt{x} - 3)} = \dfrac{4}{\sqrt{x} - 3}$ 

b. Với $x = 4$ ta có: 

 $A = \dfrac{4}{\sqrt{4} - 3} = \dfrac{4}{- 1} = - 4$ 

c. A nguyên khi $\sqrt{x} - 3$ là ước của - 4. 

Suy ra: 

$\sqrt{x} - 3 = - 4 \to \sqrt{x} = - 1$ (loại). 

$\sqrt{x} - 3 = - 2 \to \sqrt{x} = 1 \to x = 1$ 

$\sqrt{x} - 3 = - 1 \to \sqrt{x} = 2 \to x = 4$ 

$\sqrt{x} - 3 = 1 \to \sqrt{x} = 4 \to x = 16$ 

$\sqrt{x} - 3 = 2 \to \sqrt{x} = 5 \to x = 25$ 

$\sqrt{x} - 3 = 4 \to \sqrt{x} = 7 \to x = 49$