giải phương trình:
1) $(x^{2}+\frac{1}{x^2})-\frac{9}{2}(x+\frac{1}{x})+7=0$
2) $\frac{5}{x-1}+\frac{2}{-x-1}=\frac{5}{x-3}+\frac{2}{4-x}$
$2)$ Đk: $x\neq \pm 1;x\neq 3;x\neq 4$
$Pt \Rightarrow 5 \left ( \frac{1}{x-1} -\frac{1}{x-3}\right )= 2\left ( \frac{1}{4-x} +\frac{1}{x+1}\right )$
$\Leftrightarrow \frac{-10}{(x-1)(x-3)} = \frac{10}{(4-x)(x+1)}$
$\Leftrightarrow \frac{1}{-x^{2}+3x+4}+\frac{1}{x^{2}-4x+3}=0$
$\Leftrightarrow \frac{7-x}{(-x^{2}+3x+4)(x^{2}-4x+3)}=0 \Leftrightarrow x=7$ (TMĐK)