b, ĐKXĐ : \(\left\{{}\begin{matrix}x+1\ne0\\x+3\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne-1\\x\ne-3\end{matrix}\right.\)
- Ta có : \(\frac{x-3}{x+1}-\frac{x+1}{x+3}=\frac{x^2-x-10}{x^2+4x+3}\)
=> \(\frac{\left(x-3\right)\left(x+3\right)}{\left(x+1\right)\left(x+3\right)}-\frac{\left(x+1\right)\left(x+1\right)}{\left(x+3\right)\left(x+1\right)}=\frac{x^2-x-10}{\left(x+1\right)\left(x+3\right)}\)
=> \(\left(x-3\right)\left(x+3\right)-\left(x+1\right)\left(x+1\right)=x^2-x-10\)
=> \(x^2-9-x^2-2x-1-x^2+x+10=0\)
=> \(-x-x^2=0\)
=> \(x\left(x+1\right)=0\)
=> \(\left[{}\begin{matrix}x=0\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)
=> x = 0 .
Vậy phương trình có tập nghiệm là \(S=\left\{0\right\}\)
a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ne3\\x\ne-1\end{matrix}\right.\)
Ta có : \(\frac{x}{2x-6}+\frac{x}{2x+2}+\frac{2x}{\left(x+1\right)\left(3-x\right)}=0\)
=> \(\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}-\frac{4x}{\left(x+1\right)\left(x-3\right)}=0\)
=> \(x\left(x+1\right)+x\left(x-3\right)-4x=0\)
=> \(2x^2-6x=0\)
=> \(x\left(x-3\right)=0\)
=> \(\left[{}\begin{matrix}x=0\left(tm\right)\\x=3\left(ktm\right)\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{0\right\}\)
