Lời giải:

a. ĐKXĐ: $x\geq 0$

$P< \frac{1}{2}\Leftrightarrow \frac{\sqrt{x}}{\sqrt{x}+2}< \frac{1}{2}$

$\Leftrightarrow \frac{\sqrt{x}}{\sqrt{x}+2}-\frac{1}{2}<0$

$\Leftrightarrow \frac{\sqrt{x}-2}{2(\sqrt{x}+2)}<0$

$\Leftrightarrow \sqrt{x}-2<0$ (do mẫu dương rồi) 

$\Leftrightarrow 0\leq x< 4$

Kết hợp đkxđ suy ra $0\leq x< 4$

b. 

Với $x\geq 0$ thì $P\geq 0$

Lại có: $P<1$ (do tử nhỏ hơn mẫu)

$\Rightarrow P$ nguyên khi mà $P=0$

$\Leftrightarrow x=0$

cảm ơn thầy ạ

ĐKXĐ: x \(\ge\)0; x \(\ne\)1

a) P = \(\left(\frac{2}{\sqrt{x}-1}-\frac{5}{x+\sqrt{x}-2}\right):\left(1+\frac{3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right)\)

P = \(\left(\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{5}{x+2\sqrt{x}-\sqrt{x}-2}\right):\frac{x+\sqrt{x}-2+3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

P = \(\frac{2\sqrt{x}+4-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\cdot\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+1}\)

P = \(\frac{2\sqrt{x}+1}{\sqrt{x}+1}\)

b) P = \(\frac{1}{\sqrt{x}}\) <=> \(\frac{2\sqrt{x}+1}{\sqrt{x}+1}=\frac{1}{\sqrt{x}}\)

=> \(\sqrt{x}\left(2\sqrt{x}+1\right)-\sqrt{x}-1=0\)

<=> \(2x+\sqrt{x}-\sqrt{x}-1=0\)

<=> \(x=\frac{1}{2}\)(tm)

c)Với đk: x \(\ge\)0 và x \(\ne\)1

 \(x-2\sqrt{x-1}=0\) (đk: \(x\ge1\))

<=> \(x-1-2\sqrt{x-1}+1=0\)

<=> \(\left(\sqrt{x-1}-1\right)^2=0\)

<=> \(\sqrt{x-1}-1=0\)

<=> \(\sqrt{x-1}=1\)

<=> \(\left(\sqrt{x-1}\right)^2=1\)

<=> \(\left|x-1\right|=1\)

<=> \(\orbr{\begin{cases}x=0\left(ktm\right)\\x=2\left(tm\right)\end{cases}}\)

Với x = 2 => P = \(\frac{2\sqrt{2}+1}{\sqrt{2}+1}=\frac{\left(2\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\frac{4-2\sqrt{2}+\sqrt{2}-1}{2-1}=3-\sqrt{2}\)

a) P = \(\frac{2\sqrt{x}-1}{\sqrt{x}+1}\)(sửa lại)

b)  \(\frac{2\sqrt{x}-1}{\sqrt{x}+1}=\frac{1}{\sqrt{x}}\) => \(2x-\sqrt{x}-\sqrt{x}-1=0\)

<=> \(2x-2\sqrt{x}-1=0\)<=> \(2\left(x-\sqrt{x}+\frac{1}{4}\right)-\frac{3}{4}=0\)

<=>  \(2\left(\sqrt{x}-\frac{1}{2}\right)^2=\frac{3}{4}\) <=> \(\left(\sqrt{x}-\frac{1}{2}\right)^2=\frac{3}{8}\)....(tiếp tự lm)

Ta có: \(A=\dfrac{2+\sqrt{x}}{\sqrt{x}}=\dfrac{2}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}}=\dfrac{2}{\sqrt{x}}+1\)

A sẽ nguyên khi: \(2⋮\sqrt{x}\) hay \(\sqrt{x}\inƯ\left(2\right)\)

\(Ư\left(2\right)=\left\{-2;1;-1;2\right\}\) 

Mà: \(\sqrt{x}\ge0\)

Loại \(-1;-2\)

\(\Rightarrow\sqrt{x}\in\left\{2;1\right\}\)

\(\Rightarrow x\in\left\{4;1\right\}\)

Vậy A sẽ nguyên khi \(x\in\left\{1;4\right\}\)

Để A nguyên thì cănx +2 chia hết cho căn x

=>căn x thuộc Ư(2)

=>căn x=1 hoặc căn x=2

=>x=4 hoặc x=1

\(A=\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{2\sqrt{x}-1}{\sqrt{x}-1}+\frac{x-2}{x-3\sqrt{x}+2}\)

\(A=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\frac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\) \(+\frac{x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

\(A=\frac{x-4\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\frac{2x-5\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\) \(+\frac{x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

\(A=\frac{x-4\sqrt{x}+3-2x+5\sqrt{x}-2+x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

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

\(A=\frac{1}{\sqrt{x}-2}\)

vậy \(A=\frac{1}{\sqrt{x}-2}\)

A có nghĩa khi \(\sqrt{x}-2>0\)

                    \(\Leftrightarrow\sqrt{x}=2\)

                      \(\Leftrightarrow x=4\)

vậy \(x=4\) thì A có nghĩa

b) theo ý a) \(A=\frac{1}{\sqrt{x}-2}\)

theo bài ra \(A>2\) \(\Leftrightarrow\frac{1}{\sqrt{x}-2}>2\)

                                     \(\Leftrightarrow\frac{1}{\sqrt{x}-2}-2>0\)

                                      \(\Leftrightarrow\frac{1}{\sqrt{x}-2}-\frac{2\left(\sqrt{x}-2\right)}{\sqrt{x}-2}>0\)

                                      \(\Leftrightarrow\frac{1-2\sqrt{x}+4}{\sqrt{x}-2}>0\)

                                      \(\Leftrightarrow\frac{5-2\sqrt{x}}{\sqrt{x}-2}>0\)

\(\Rightarrow\hept{\begin{cases}5-2\sqrt{x}>0\\\sqrt{x}-2>0\end{cases}}\)  hoặc \(\hept{\begin{cases}5-2\sqrt{x}< 0\\\sqrt{x}-2< 0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}-2\sqrt{x}>-5\\\sqrt{x}>2\end{cases}}\) hoặc \(\hept{\begin{cases}-2\sqrt{x}< -5\\\sqrt{x}< 2\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x< \frac{25}{4}\\x>4\end{cases}}\)hoặc \(\hept{\begin{cases}x>\frac{25}{4}\\x< 4\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}4< x< \frac{25}{4}\\x\notin\varnothing\end{cases}}\)

vậy \(4< x< \frac{25}{4}\) thì \(A>2\)

mình sửa lại chút chỗ dòng thứ 2 từ dưới lên

\(\Rightarrow\orbr{\begin{cases}4< x< \frac{25}{4}\\x\in\varnothing\end{cases}}\)

mải quá nên mình ấn mhầm cho mk xin lỗi

diều kiện x >= 0

P=\(\left(\frac{x+2}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}\right).\frac{4\sqrt{x}}{3}\)

= \(\frac{x+2-x+\sqrt{x}-1}{x\sqrt{x}+1}.\frac{4\sqrt{x}}{3}\)

=\(\frac{\sqrt{x}+1}{x\sqrt{x}+1}.\frac{4\sqrt{x}}{3}\)=\(\frac{4\sqrt{x}}{3x-3\sqrt{x}+3}\)

P=8/9

<=> \(\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\frac{8}{9}\)

<=> \(3\sqrt{x}=2x-2\sqrt{x}+1\)

<=> \(2x-5\sqrt{x}+2=0\)

<=> \(\left[\begin{array}{nghiempt}x=4\\x=\frac{1}{4}\end{array}\right.\)

vậy x=4 hoặc x=1/4 thì p=8/9

a) \(P=\left(\frac{x+2}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}\right)\cdot\frac{4\sqrt{x}}{3}\left(ĐK:x\ge0;x\ne-1\right)\)

\(=\left[\frac{x+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}+1}\right]\cdot\frac{4\sqrt{x}}{3}\)

\(=\frac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\frac{4\sqrt{x}}{3}\)

\(=\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\frac{4\sqrt{x}}{3}\)

\(=\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)

b) Để P=8/9

\(\Leftrightarrow\)\(\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\frac{8}{9}\)

\(\Leftrightarrow24\left(x-\sqrt{x}+1\right)=36\sqrt{x}\)

\(\Leftrightarrow24x-24\sqrt{x}+24-36\sqrt{x}=0\)

\(\Leftrightarrow24x-60\sqrt{x}+24=0\)

\(\Leftrightarrow12\left(2x-5\sqrt{x}+2\right)=0\)

\(\Leftrightarrow\left(2x-\sqrt{x}\right)-\left(4\sqrt{x}-2\right)=0\)

\(\Leftrightarrow\sqrt{x}\left(2\sqrt{x}-1\right)-2\left(2\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}2\sqrt{x}-1=0\\\sqrt{x}-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}\sqrt{x}=\frac{1}{2}\\\sqrt{x}=2\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{1}{4}\left(tm\right)\\x=4\left(tm\right)\end{array}\right.\)

help me !!!!!!!!!!!!!!