Question

\(\frac{x+3}{x-3}+\frac{48}{9-x^2}=\frac{x-3}{x+3}\)

Answer 1

\(\frac{x+3}{x-3}+\frac{48}{9-x^2}=\frac{x-3}{x+3}\)

ĐKXĐ: \(x\ne3;x\ne-3\)

\(\Leftrightarrow\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}-\frac{48}{\left(x-3\right)\left(x+3\right)}-\frac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Rightarrow x^2+6x+9-48-x^2+6x-9=0\)

\(\Leftrightarrow12x-48=0\)

\(\Leftrightarrow12x=48\)

\(\Leftrightarrow x=4\left(TM\right)\)

Vậy phương trình có tập nghiệm là: S ={4}