Bạn nên gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề và hỗ trợ bạn tốt hơn nhé.

Ta có: P = x − x + 2 ( x + 1 ) ( x − 2 ) − x x ( x − 2 ) : 1 − x 2 − x = x − x + 2 − x ( x + 1 ) ( x + 1 ) ( x − 2 ) . 2 − x 1 − x = 2 − 2 x ( x + 1 ) ( x − 1 ) = 2 ( 1 − x ) ( x + 1 ) ( x − 1 ) = − 2 x + 1  

?

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

\(=\dfrac{x-4}{x}\cdot\dfrac{x-3\sqrt{x}+2-x-3\sqrt{x}-2}{x-4}=\dfrac{-6}{\sqrt{x}}\)

a: Khi x=64 thì \(A=\dfrac{2}{8-2}=\dfrac{2}{6}=\dfrac{1}{3}\)

b: \(P=B:A\)

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

\(=\dfrac{4\sqrt{x}-2-2\sqrt{x}-4}{x-4}\cdot\dfrac{\sqrt{x}-2}{2}\)

\(=\dfrac{2\sqrt{x}-6}{2\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\)

c: P<0

=>căn x-3<0

=>0<=x<9

mà x nguyên và x<>4

nên \(x\in\left\{0;1;2;3;5;6;7;8\right\}\)

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

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

1. \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{4}{x+2\sqrt{x}}\right):\left(1+\dfrac{1}{\sqrt{x}}\right)\left(x>0\right)\)

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

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

2. Để \(P>0\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}>0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}-2>0\\\sqrt{x}+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}-2< 0\\\sqrt{x}+1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\sqrt{2}\\x>\sqrt{-1}\left(L\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x< \sqrt{2}\\x< \sqrt{-1}\left(L\right)\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x>\sqrt{2}\\x< \sqrt{2}\end{matrix}\right.\)

Vậy \(P>0\Leftrightarrow\left[{}\begin{matrix}x>\sqrt{2}\\x< \sqrt{2}\end{matrix}\right.\)