\(\frac{\left(x+3\right)x}{x\left(x+1\right)}+\frac{\left(x+1\right)\left(x-3\right)}{x\left(x+1\right)}=\frac{2x\left(x+1\right)}{x\left(x+1\right)}\)
x2+3x+x2+x-x+3=2x2+2x
x2+x2-2x2+3x+x-x-2x=-3
x=-3
\(\frac{x+3}{x+1}+\frac{x-3}{x}=2\)
\(\left(x+3\right)x+\left(x-3\right)\left(x-1\right)=2\left(x+1\right)x\)
\(x^2+3x+x^2-4x+3=2x^2-2x\)
\(x^2+x^2-2x^2+3x-4x+2x+3=0\)
\(x+3=0\Rightarrow x=-3\)
\(\frac{x+3}{x+1}+\frac{x-3}{x}=2\)
\(x\left(x+3\right)+\left(x-3\right)\left(x+1\right)=2x\left(x+1\right)\)
\(2x^2+x-3=2x^2+2x\)
\(2x^2+x-3-2x^2-2x=0\)
\(-1x-3=0\)
\(x=-3\)
\(\frac{x+3}{x+1}\)+\(\frac{x-3}{x}\)=2 (ĐKXĐ:x≠-1,x≠0)
⇔\(\frac{x\left(x+3\right)}{x\left(x+1\right)}\)+\(\frac{\left(x-3\right)\left(x+1\right)}{x\left(x+1\right)}\)=\(\frac{2x\left(x+1\right)}{x\left(x+1\right)}\)
⇒x2+3x+x2+x-3x-3=2x2+2x
⇔-x-3=0
⇔x=-3 (TMĐK)
Vậy tập nghiệm của phương trình đã cho là:S={-3}