\(a,\frac{1}{3-x}-\frac{1}{x+1}=\frac{x}{x-3}-\frac{\left(x-1\right)^2}{x^2-2x-3}\)\(Đkxđ:\left\{{}\begin{matrix}x\ne-1\\x\ne3\end{matrix}\right.\)
\(\Leftrightarrow\frac{1}{3-x}-\frac{1}{x+1}=\frac{x}{x-3}-\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-3\right)}\)
\(\Leftrightarrow\frac{x}{x-3}-\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-3\right)}+\frac{1}{x+1}+\frac{1}{x-3}=0\)
\(\Leftrightarrow\frac{x\left(x+1\right)-\left(x-1\right)^2+\left(x-3\right)+\left(x+1\right)}{\left(x+1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow x^2+x-x^2+2x-1+x-3+x+1=0\)
\(\Leftrightarrow5x-3=0\)
\(\Leftrightarrow x=\frac{3}{5}\left(tmđk\right)\)
Vậy ......
\(b,\frac{2}{x+2}-\frac{2x^2+16}{x^3+8}=\frac{5}{x^2-2x+4}\) \(Đkxđ:....\)
\(\Leftrightarrow\frac{2\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}-\frac{2x^2+16}{\left(x+2\right)\left(x^2-2x+4\right)}=\frac{5\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\)
\(\Leftrightarrow-4x+8-16=10\)
\(\Leftrightarrow x=-\frac{9}{2}\)
Vậy ...............
