Với `x \ne +-2` có:

`M=[x^3]/[x^2-4]-x/[x-2]-2/[x+2]`

`M=[x^3-x(x+2)-2(x-2)]/[(x-2)(x+2)]`

`M=[x^3-x^2-2x-2x+4]/[(x-2)(x+2)]`

`M=[x^3-x^2-4x+4]/[(x-2)(x+2)]`

`M=[x^2(x-1)-4(x-1)]/[x^2-4]`

`M=[(x-1)(x^2-4)]/[x^2-4]`

`M=x-1`

\(M=\dfrac{x^3}{x^2-4}-\dfrac{x}{x-2}-\dfrac{2}{x+2}\)

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

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

\(=\dfrac{\left(x^2-4\right)\left(x-1\right)}{x^2-4}=x-1\)

Với  có:

\(a,x\ne\pm2\\ b,\\ =\dfrac{x^3-x\left(x+2\right)-2\left(x-2\right)}{x^2-4}\\ =\dfrac{x^3-x^2-2x-2x+4}{x^2-4}=\dfrac{x^3-4x-x^2+4}{x^2-4}\\ =\dfrac{x\left(x^2-4\right)-\left(x^2-4\right)}{x^2-4}=\dfrac{\left(x^2-4\right)\left(x-1\right)}{x^2-4}\\ =x-1\)

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

\(\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^3-2x^2-4x+8}\)

Để biểu thức trên nhận giá trị âm khi \(\dfrac{\left(x-2\right)^2}{x^3-2x^2-4x+8}< 0\)

\(\Rightarrow x^3-2x^2-4x+8< 0\)do \(\left(x-2\right)^2\ge0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-2x+4\right)-2x\left(x+2\right)< 0\)

\(\Leftrightarrow\left(x+2\right)\left(x-2\right)^2< 0\Leftrightarrow x< -2\)

a.\(A=\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{\left(x^2-4\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)

\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)

(-3).8/8.6 rút gọn

Bài 1: 

a: \(A=\dfrac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}=\dfrac{x\left(x+1\right)}{x\left(x-3\right)}=\dfrac{x+1}{x-3}\)

b: Để A=3 thì 3x-9=x+1

=>2x=10

hay x=5

Bài 2: 

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

\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{2}=\dfrac{-3}{x-2}\)

b: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)

hay \(x\in\left\{3;1;5;-1\right\}\)

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.