Question

giải pt sau

x+4/x+1+x/x-1=2x^2/x^2-1

Answer 1

\(\dfrac{x+4}{x+1}+\dfrac{x}{x-1}=\dfrac{2x^2}{x^2-1}\) ĐKXĐ: \(x\ne1;x\ne-1\)

\(\Leftrightarrow\dfrac{\left(x+4\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\dfrac{x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{2x^2}{\left(x+1\right)\left(x-1\right)}\)

\(\Rightarrow x^2+3x-4+x^2+1=2x^2\)

\(\Leftrightarrow x^2+x^2-2x^2+3x=4-1\)

\(\Leftrightarrow3x=3\)

\(\Leftrightarrow x=1\)

Answer 2

vậy pt trên vô nghiem