Lời giải:
ĐKXĐ: $x\neq \pm 4$
PT \(\Leftrightarrow \frac{x^2-8}{x^2-16}=\frac{x-4+x+4}{(x-4)(x+4)}=\frac{2x}{x^2-16}\)
\(\Rightarrow x^2-8=2x\)
\(\Leftrightarrow x^2-2x-8=0\)
\(\Leftrightarrow (x-1)^2-9=0\Leftrightarrow (x-1)^2-3^2=0\)
\(\Leftrightarrow (x-4)(x+2)=0\Rightarrow \left[\begin{matrix} x=4\\ x=-2\end{matrix}\right.\)
\(\frac{x^2-8}{x^2-16}=\frac{1}{x+4}+\frac{1}{x-4}\)
\(\Leftrightarrow\frac{x^2-8}{\left(x-4\right)\left(x+4\right)}-\frac{\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}-\frac{x+4}{\left(x-4\right)\left(x+4\right)}=0\)
\(\Leftrightarrow\)\(\frac{x^2-8-x+4-x-4}{\left(x-4\right)\left(x+4\right)}=0\)
\(\Leftrightarrow\frac{x^2-2x-8}{\left(x-4\right)\left(x+4\right)}=0\)
\(\Leftrightarrow\frac{\left(x-4\right)\left(x+2\right)}{\left(x-4\right)\left(x+4\right)}=0\)
\(\Leftrightarrow\)\(\frac{x+2}{x+4}=0\)
\(\Leftrightarrow x+2=0\Rightarrow x=-2\)
Giải phương trình sau :
\(\frac{x^2-8}{x^2-16}=\frac{1}{x+4}+\frac{1}{x-4}\)
⇔\(\frac{x^2-8}{x^2-16}=\frac{1\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}+\frac{1\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}\)
➞\(x^2-8\)=x -4+ x+4
⇔x2-8=2x
⇔x2-2x=8
⇔ x(x-2)=8
⇔ x=8 hoặc x-2=8
⇔x=8 hoặc x= 10
