a) ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)

Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right):\dfrac{2\sqrt{x}}{x-4}\)

\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\cdot\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right):\dfrac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

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

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

\(=\sqrt{x}\)

b) Để P>4 thì \(\sqrt{x}>4\)

hay x>16

Kết hợp ĐKXĐ, ta được: x>16

Vậy: Khi x>16 thì P>4

\(=2\sqrt{3}-4\sqrt{3}+5\sqrt{3}=3\sqrt{3}\)

Bài 5: 

\(\widehat{B}=60^0\)

\(AB=8\sqrt{3}\left(cm\right)\)

\(BC=16\sqrt{3}\left(cm\right)\)

a) 

ĐKXĐ: \(x-4\ge0\text{ (1)};\text{ }x+4\sqrt{x-4}\ge0\text{ (2); }\frac{16}{x^2}-\frac{8}{x}+1>0\text{ (3)}\)

\(\left(1\right)\Leftrightarrow x\ge4\)

\(\left(2\right)\Leftrightarrow\left(\sqrt{x-4}+2\right)^2\ge0\text{ (đúng }\forall x\ge4\text{)}\)

\(\left(3\right)\Leftrightarrow\left(\frac{4}{x}-1\right)^2>0\Leftrightarrow\frac{4}{x}-1\ne0\Leftrightarrow x\ne4\)

Vậy ĐKXĐ là \(x>4\)

b)

\(A=\frac{\left|\sqrt{x-4}+2\right|+\left|\sqrt{x-4}-2\right|}{\left|\frac{4}{x}-1\right|}=\frac{\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|}{1-\frac{4}{x}}=\frac{x\left(\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|\right)}{x-4}\)

\(+\sqrt{x-4}\le2\Leftrightarrow0<\)\(x-4\le4\)

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

A nguyên khi \(\frac{16}{x-4}\)nguyên hay \(x-4\inƯ\left(16\right)\)

Mà \(0<\)\(x-4\le4\)

Nên \(x-4\in\left\{2;4\right\}\Rightarrow x\in\left\{6;8\right\}\)

\(+\text{Xét }\sqrt{x-4}>2\Leftrightarrow x-4>4\)

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

Nếu \(\sqrt{x-4}\)là số vô tỉ thì A là số vô tỉ.

Để A là hữu tỉ thì \(\sqrt{x-4}=t\text{ }\left(t\in Z;\text{ }t>4\right)\Rightarrow x=t^2+4\)

Khi đó, \(A=\frac{2\left(t^2+4\right)}{t}=2t+\frac{8}{t}\)

A nguyên khi \(\frac{8}{t}\) nguyên hay \(t=8\text{ (do }t>4\text{)}\)

\(t=\sqrt{x-4}=8\Leftrightarrow x=8^2+4=68\)

Vậy \(x\in\left\{6;8;68\right\}\)

c/

\(+0<\sqrt{x-4}\)\(<2\)

Thì \(A=4+\frac{16}{x-4}>4+\frac{16}{4}=8\)

\(+\sqrt{x-4}\ge2\)

\(A=\frac{2x}{\sqrt{x-4}}=2t+\frac{8}{t}\text{ (}t=\sqrt{x-4}\ge2\text{)}\)

 Mà \(t+\frac{4}{t}\ge2\sqrt{t.\frac{4}{t}}=4\)

\(\Rightarrow A\ge2.4=8\)

Dấu "=" xảy ra khi \(t=\frac{4}{t}\Leftrightarrow t=2\Leftrightarrow\sqrt{x-4}=2\Leftrightarrow x=8\)

Vậy GTNN của A là 8 khi x = 8.

a: ĐKXĐ: x>0; x<>4

b: \(A=\dfrac{4+4\sqrt{x}+x-4+4\sqrt{x}-x+4x}{4-x}:\dfrac{2\sqrt{x}-\sqrt{x}-3}{2\sqrt{x}-x}\)

\(=\dfrac{4\sqrt{x}\left(\sqrt{x}+2\right)}{4-x}\cdot\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}=\dfrac{4x}{\sqrt{x}-3}\)

c: \(A-1=\dfrac{4x-\sqrt{x}+3}{\sqrt{x}-3}< 0\)

=>căn x-3<0

=>0<x<9 và x<>4

AI GIẢI HỘ MÌNH K CHO Ạ!!!

1)  a) Căn thức có nghĩa \(\Leftrightarrow4-2x\ge0\Leftrightarrow2x\le4\Leftrightarrow x\le2\)

b) Thay x = 2 vào biểu thức A, ta được: \(A=\sqrt{4-2.2}=\sqrt{0}=0\)

Thay x = 0 vào biểu thức A, ta được: \(A=\sqrt{4-2.0}=\sqrt{4}=2\)

Thay x = 1 vào biểu thức A, ta được: \(A=\sqrt{4-2.1}=\sqrt{2}\)

Thay x = -6 vào biểu thức A, ta được: \(A=\sqrt{4-2.\left(-6\right)}=\sqrt{16}=4\)

Thay x = -10 vào biểu thức A, ta được: \(A=\sqrt{4-2.\left(-10\right)}=\sqrt{24}=2\sqrt{6}\)

c) \(A=0\Leftrightarrow\sqrt{4-2x}=0\Leftrightarrow4-2x=0\Leftrightarrow x=2\)

\(A=5\Leftrightarrow\sqrt{4-2x}=5\Leftrightarrow4-2x=25\Leftrightarrow x=\frac{-21}{2}\)

\(A=10\Leftrightarrow\sqrt{4-2x}=10\Leftrightarrow4-2x=100\Leftrightarrow x=-48\)

a) Để \(\sqrt{\left|x\right|-1}\) xác định 

<=> \(\left|x\right|\ge1\)

<=> \(\left[{}\begin{matrix}x\ge1\\x\le-1\end{matrix}\right.\)

b) Để \(\sqrt{-\left|x+5\right|}\) xác định

<=> \(-\left|x+5\right|\ge0\)

Mà \(\left|x+5\right|\ge0\left(\forall x\right)\)

<=> x + 5 = 0 <=> x = -5

c) Để \(\sqrt{\left|x-1\right|-3}\) xác định

<=> \(\left|x-1\right|\ge3\)

<=> \(\left[{}\begin{matrix}x-1\ge3< =>x\ge4\\x-1\le-3< =>x\le-2\end{matrix}\right.\)

`a)đk:|x|-1>=0`

`<=>|x|>=1`

`<=>` \(\left[ \begin{array}{l}x \ge 1\\x\le -1\end{array} \right.\) 

`b)đk:-|x+5|>=0`

`<=>|x+5|<=0`

Mà `|x+5|>=0`

`<=>|x+5|=0`

`<=>x=-5`

`c)đk:|x-1|-3>=0`

`|x-1|>=3`

`<=>` \(\left[ \begin{array}{l}x-1 \ge 3\\x-1 \le -3\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x \ge 4\\x \le -2\end{array} \right.\) 

\(a,ĐK:x>0;x\ne9\\ b,A=\dfrac{\sqrt{x}+3+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\\ A=\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}=\dfrac{2}{\sqrt{x}+3}\\ c,A>\dfrac{2}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+3}-\dfrac{2}{5}>0\\ \Leftrightarrow\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{5}>0\\ \Leftrightarrow\dfrac{2-\sqrt{x}}{5\left(\sqrt{x}+3\right)}>0\\ \Leftrightarrow2-\sqrt{x}>0\left(\sqrt{x}+3>0\right)\\ \Leftrightarrow\sqrt{x}< 2\Leftrightarrow0< x< 4\)