\(\dfrac{2x}{x^2-1}-\dfrac{1}{x+1}=2\) ĐKXĐ: \(x\ne\pm1\)
\(\Leftrightarrow\dfrac{2x}{\left(x+1\right)\left(x-1\right)}-\dfrac{x-1}{\left(x+1\right)\left(x-1\right)}=\dfrac{2\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(\Leftrightarrow\dfrac{2x-x+1}{\left(x+1\right)\left(x-1\right)}=\dfrac{2x^2-2}{\left(x+1\right)\left(x-1\right)}\)
\(\Rightarrow x+1=2x^2-2\)
\(\Leftrightarrow2x^2-x-3=0\)
có \(a-b+c=2+1-3=0\)
\(\Rightarrow x_1=-1\)(loại)
\(x_2=\dfrac{3}{2}\)(nhận)
vậy phương trình đã cho có một nghiệm x=3/2