ĐKXĐ : \(x\ne\pm2;x\ne0;x\ne3\)

\(A=\left(\frac{4x}{2+x}+\frac{8x^2}{4-x^2}\right):\left(\frac{x-1}{x^2-2x}-\frac{2}{x}\right)\)

\(=\frac{4x\left(2-x\right)+8x^2}{\left(2-x\right)\left(2+x\right)}:\frac{x-1-2\left(x-2\right)}{x\left(x-2\right)}\)

\(=\frac{8x-4x^2+8x^2}{\left(2-x\right)\left(2+x\right)}:\frac{x-1-2x+4}{x\left(x-2\right)}\)

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

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

\(=\frac{4x^2}{x-3}\)

\(A< 0\Leftrightarrow\frac{4x^2}{x-3}< 0\Leftrightarrow x-3< 0\) ( do \(4x^2>0\) )

\(\Leftrightarrow x< 3\) 

Vậy :........

Ta có: \(B=\left(\frac{4x}{x+2}+\frac{8x^2}{4-x^2}\right):\left(\frac{x-1}{x^2-2x}-\frac{2}{x}\right)\)

\(=\left(\frac{4x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{8x^2}{\left(x-2\right)\left(x+2\right)}\right):\left(\frac{x-1}{x\left(x-2\right)}-\frac{2\left(x-2\right)}{x\left(x-2\right)}\right)\)

\(=\frac{4x^2-8x-8x^2}{\left(x-2\right)\left(x+2\right)}:\frac{x-1-2x+4}{\left(x-2\right)}\)

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

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

\(=-\frac{4x}{3-x}=\frac{4x}{x-3}\)

a) ĐKXĐ: \(x\notin\left\{2;-2;0;3\right\}\)

Để B=-1 thì \(\frac{4x}{x-3}=-1\)

\(\Leftrightarrow4x=3-x\)

\(\Leftrightarrow4x+x=3\)

\(\Leftrightarrow5x=3\)

hay \(x=\frac{3}{5}\)(nhận)

Vậy: Để B=-1 thì \(x=\frac{3}{5}\)

b) Sửa đề: Tìm x để B<0

Để B<0 thì \(\frac{4x}{x-3}< 0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4x>0\\x-3< 0\end{matrix}\right.\\\left\{{}\begin{matrix}4x< 0\\x-3>0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x< 3\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x>3\end{matrix}\right.\end{matrix}\right.\Leftrightarrow0< x< 3\)

Kết hợp ĐKXĐ, ta được:

\(\left\{{}\begin{matrix}0< x< 3\\x\ne2\end{matrix}\right.\)

Vậy: Để B<0 thì \(\left\{{}\begin{matrix}0< x< 3\\x\ne2\end{matrix}\right.\)

Hình như đề sai.Sửa đề luôn nha !

\(ĐKXĐ:x\ne\pm2\)

\(A=\left(\frac{x}{x^2-4}-\frac{2}{x-2}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)

\(=\left(\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right):\frac{\left(x-2\right)\left(x+2\right)+10-x^2}{x+2}\)

\(=\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\frac{x^2-4+10-x^2}{x+2}\)

\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}\cdot\frac{x+2}{-6}=\frac{1}{x-2}\)

b

Để \(A< 0\Rightarrow\frac{1}{x-2}< 0\Rightarrow x-2< 0\Rightarrow x< 2\)

c

Để A nguyên thì \(\frac{1}{x-2}\) nguyên

\(\Rightarrow1⋮x-2\)

\(\Rightarrow x-2\in\left\{1;-1\right\}\Rightarrow x\in\left\{3;1\right\}\)