a: ĐKXĐ: x∉{0;-2;1;-1}
\(D=\left(\frac{x}{x+2}+\frac{8x+8}{x^2+2x}-\frac{x+2}{x}\right):\left(\frac{x^2-x-3}{x^2+2x}+\frac{1}{x}\right)\)
\(=\left(\frac{x}{x+2}+\frac{8x+8}{x\left(x+2\right)}-\frac{x+2}{x}\right):\left(\frac{x^2-x-3}{x\left(x+2\right)}+\frac{1}{x}\right)\)
\(=\frac{x^2+8x+8-\left(x+2\right)^2}{x\left(x+2\right)}:\frac{x^2-x-3+x+2}{x\left(x+2\right)}\)
\(=\frac{x^2+8x+8-x^2-4x-4}{x\left(x+2\right)}\cdot\frac{x\left(x+2\right)}{x^2-1}\)
\(=\frac{4x+4}{x^2-1}=\frac{4\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{4}{x-1}\)
b: x(x-2)-x+2=0
=>x(x-2)-(x-2)=0
=>(x-2)(x-1)=0
=>\(\left[\begin{array}{l}x-2=0\\ x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\left(nhận\right)\\ x=1\left(loại\right)\end{array}\right.\)
Khi x=2 thì \(D=\frac{4}{2-1}=\frac41=4\)
c: D<0
=>\(\frac{4}{x-1}<0\)
=>x-1<0
=>x<1
Kết hợp ĐKXĐ, ta được:
\(\begin{cases}x<1\\ x\notin\left\lbrace0;-2;-1\right\rbrace\end{cases}\)
