Question

Answer 1

a)\(A=\dfrac{x\left(\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(2-\sqrt{x-4}\right)^2}\right)}{\sqrt{\left(x-4\right)^2}}\)

\(=\dfrac{x\left(2+2\right)}{x-4}\)

\(=\dfrac{4x}{x-4}=4+\dfrac{16}{x-4}\)

Để A có giá trị nguyên

\(\Rightarrow x-4\in U\left(16\right)=\left\{1,-1,2,-2,4,-4,8,-8,16,-16\right\}\)

\(\Rightarrow x\in\left\{5,3,6,2,0,8,12,-4,20,-12\right\}\)

 

 

 

Answer 2

\(a,x>4\Leftrightarrow x-4>0\Leftrightarrow\left|x-4\right|=x-4\\ A=\dfrac{x\left[\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}\right]}{\sqrt{\left(x-4\right)^2}}\\ A=\dfrac{x\left(\sqrt{x-4}+2+\sqrt{x-4}-2\right)}{x-4}\\ A=\dfrac{2x\sqrt{x-4}}{x-4}=\dfrac{2x}{\sqrt{x-4}}=\dfrac{2x-8+8}{\sqrt{x-4}}\\ A=\dfrac{2\left(x-4\right)+8}{\sqrt{x-4}}=2\sqrt{x-4}+\dfrac{8}{\sqrt{x-4}}\\ \ge2\sqrt{2\sqrt{x-4}\cdot\dfrac{8}{\sqrt{x-4}}}=2\cdot4=8\left(BĐT.cosi\right)\\ A_{min}=8\Leftrightarrow2\left(x-4\right)=8\Leftrightarrow x=8\left(tm\right)\)

\(b,A\in Z\Leftrightarrow\dfrac{2x}{\sqrt{x-4}}\in Z\Leftrightarrow2\sqrt{x-4}+\dfrac{8}{\sqrt{x-4}}\in Z\left(câu.a\right)\\ \Leftrightarrow8⋮\sqrt{x-4}\\ \Leftrightarrow\sqrt{x-4}\inƯ\left(8\right)=\left\{1;2;4;8\right\}\left(\sqrt{x-4}>0\right)\\ \Leftrightarrow x-4\in\left\{1;4;16;64\right\}\\ \Leftrightarrow x\in\left\{5;8;20;68\right\}\left(tm\right)\)

Answer 3

Lời giải:
a. \(A=\frac{x[\sqrt{(\sqrt{x-4}+2)^2}+\sqrt{(\sqrt{x-4}-2)^2}]}{\sqrt{(x-4)^2}}=\frac{x(|\sqrt{x-4}+2|+|\sqrt{x-4}-2|)}{|x-4|}\)

TH1: $x\geq 8$ thì: \(A=\frac{x(\sqrt{x-4}+2+\sqrt{x-4}-2)}{x-4}=\frac{2x\sqrt{x-4}}{x-4}=\frac{2x}{\sqrt{x-4}}\)

\(A=\frac{2(x-4)+8}{\sqrt{x-4}}\geq \frac{2\sqrt{16(x-4)}}{\sqrt{x-4}}=8\)

Vậy $A_{\min}=8$ khi $x=8$

TH2: $4< x< 8$ thì:

\(A=\frac{x(\sqrt{x-4}+2+2-\sqrt{x-4})}{x-4}=\frac{4x}{x-4}=\frac{4(x-4)+16}{x-4}=4+\frac{16}{x-4}\)

Vì $x< 8$ nên $A> 4+\frac{16}{8-4}=8$

Vậy từ 2 TH ta có $A_{\min}=8$ khi $x=8$. 

b. TH1: $x\geq 8$.

Với $x$ nguyên thì để $A=\frac{2x}{\sqrt{x-4}}$ nguyên thì trước tiên $\sqrt{x-4}$ phải là số nguyên.

Đặt $\sqrt{x-4}=t(t\geq 2)$ và $t$ nguyên thì:
$A=\frac{2(t^2+4)}{t}=2t+\frac{8}{t}$

$A\in\mathbb{Z}\Leftrightarrow \frac{8}{t}\in\mathbb{Z}\Leftrightarrow t=2,4,8$ (do $t\geq 2$)

Thay vô thì $A$ nhận giá trị nguyên khi $x=8; 20, 68$

TH2: $4< x< 8$ thì vì $x$ nguyên nên $x=5;6;7$. Thay vô ta thấy $x=5,6$ thỏa mãn.

Vậy.......