Question

giải phương trình : \(\dfrac{x-4}{x-1}\)+ \(\dfrac{x+4}{x+1}\) = 2

Answer 1

ĐKXĐ:\(x\ne\pm1\)

\(\dfrac{x-4}{x-1}+\dfrac{x+4}{x+1}=2\\ \Leftrightarrow\dfrac{\left(x+1\right)\left(x-4\right)}{\left(x+1\right)\left(x-1\right)}+\dfrac{\left(x-1\right)\left(x+4\right)}{\left(x+1\right)\left(x-1\right)}=2\\ \Leftrightarrow\dfrac{x^2-3x-4}{\left(x+1\right)\left(x-1\right)}+\dfrac{x^2+3x-4}{\left(x+1\right)\left(x-1\right)}=2\\ \Leftrightarrow\dfrac{x^2-3x-4+x^2+3x-4}{x^2-1}=2\\ \Rightarrow2x^2-8=2x^2-2\\ \Leftrightarrow-8=-2\left(vô.lí\right)\)