Lời giải:
a) ĐKXĐ: \(x\neq -1; -2\)
PT \(\Leftrightarrow (x^2+2)\left(\frac{1}{x^2-2x+2}-\frac{1}{x^2+3x+2}\right)=\frac{5}{2}\)
\(\Leftrightarrow (x^2+2).\frac{5x}{(x^2-2x+2)(x^2+3x+2)}=\frac{5}{2}\)
\(\Leftrightarrow 2x(x^2+2)=(x^2-2x+2)(x^2+3x+2)\)
Đặt \(x^2+2=a\) thì pt trở thành:
\(2xa=(a-2x)(a+3x)\)
\(\Leftrightarrow a^2-ax-6x^2=0\)\(\Leftrightarrow (a-3x)(a+2x)=0\Rightarrow \left[\begin{matrix} a-3x=0\\ a+2x=0\end{matrix}\right.\)
\(\Leftrightarrow \left[\begin{matrix} x^2-3x+2=0\\ x^2+2+2x=0\end{matrix}\right.\Rightarrow x=1; x=2\)
Vậy..........
b) ĐK: $x\neq -2$
PT \(\Leftrightarrow (x+2)^2+\frac{4x^2}{(x+2)^2}-4x=9\)
\(\Leftrightarrow (x+2)^2+(\frac{2x}{x+2})^2-2.(x+2).\frac{2x}{x+2}=9\)
\(\Leftrightarrow (x+2-\frac{2x}{x+2})^2=9\)
\(\Rightarrow \left[\begin{matrix} x+2-\frac{2x}{x+2}=3\\ x+2-\frac{2x}{x+2}=-3\end{matrix}\right.\Rightarrow \left[\begin{matrix} \frac{x^2}{x+2}=1\\ \frac{x^2}{x+2}=-5\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x^2-x-2=0\\ x^2+5x+10=0\end{matrix}\right.\Rightarrow x=2; x=-1\) (thỏa mãn)
Vậy..........
