\(\dfrac{x}{x+4}+\dfrac{56-x^2}{x^2-16}=\dfrac{5}{x-4}\) (*)
ĐKXĐ: \(x\ne4\)
(*) \(\Leftrightarrow\dfrac{x^2-4x}{\left(x-4\right)\left(x+4\right)}+\dfrac{56-x^2}{\left(x-4\right)\left(x+4\right)}=\dfrac{5x+20}{\left(x-4\right)\left(x+4\right)}\)
\(\Leftrightarrow\dfrac{56-4x}{\left(x-4\right)\left(x+4\right)}=\dfrac{5x+20}{\left(x-4\right)\left(x+4\right)}\)
\(\Leftrightarrow56-4x=5x+20\)
\(\Leftrightarrow9x=36\)
\(\Leftrightarrow x=4\) ( không thỏa mãn đk )
Vậy phương trình vô nghiệm
b: Ta có: \(\dfrac{x}{x+4}+\dfrac{56-x^2}{x^2-16}=\dfrac{5}{x-4}\)
Suy ra: \(x^2-4x+56-x^2=5x+20\)
\(\Leftrightarrow5x+20=-4x+56\)
\(\Leftrightarrow9x=36\)
hay x=4(loại)