ĐKXĐ: \(x\notin\left\{0;-1\right\}\)
\(\dfrac{x+a}{x+1}+\dfrac{x-2}{x}=2\)
=>\(\dfrac{x\cdot\left(x+a\right)+\left(x-2\right)\left(x+1\right)}{x\left(x+1\right)}=\dfrac{2x\left(x+1\right)}{x\left(x+1\right)}\)
=>\(x^2+a\cdot x+x^2-x-2=2x^2+2x\)
=>\(ax-x-2=2x\)
=>ax-x-2x=2
=>ax-3x=2
=>x(a-3)=2
Để phương trình vô nghiệm thì a-3=0
=>a=3

