\(\frac{2x}{x-3}+\frac{x+1}{x+3}+\frac{4}{9-x^2}=0\)
ĐKXĐ: \(x\ne3;x\ne-3\)
\(\Leftrightarrow\frac{2x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x+1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{4}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Rightarrow2x^2+6x+x^2-3x+x-3-4=0\)
\(\Leftrightarrow3x^2+7x-3x-7=0\)
\(\Leftrightarrow\left(3x^2-3x\right)+\left(7x-7\right)=0\)
\(\Leftrightarrow3x\left(x-1\right)+7\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\3x+7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(TM\right)\\x=-\frac{7}{3}\left(TM\right)\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là: \(S=\left\{1;-\frac{7}{3}\right\}\)
\(\frac{2x}{x-3}+\frac{x+1}{x+3}+\frac{4}{9-x^2}=0\)
ĐKXĐ : \(x\ne\pm3\)
\(\Leftrightarrow2x\left(x+3\right)+\left(x+1\right)\left(x-3\right)-4=0\)
\(\Leftrightarrow2x^2+6x+x^2-3x+x-3-4=0\)
\(\Leftrightarrow3x^2+4x-7=0\)
\(\Leftrightarrow3x^2+7x-3x-7=0\)
\(\Leftrightarrow3x\left(x-1\right)+7\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(3x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\3x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{-7}{3}\end{matrix}\right.\left(nhận\right)\)
Vậy : \(S=\left\{\frac{-7}{3};1\right\}\)