Question

\(\dfrac{x+1}{x-1}\)-\(\dfrac{1}{x+1}\)=\(\dfrac{x^2+2}{x^2-1}\)

giúp mik vs ah

 

Answer 1

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

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

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

\(\Leftrightarrow\left(x+1\right)\left(x+1\right)-\left(x-1\right)=x^2+2\)

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

\(\Leftrightarrow x^2+2x+1-x+1-x^2-2=0\)

\(\Leftrightarrow x=0\left(tm\right)\)

Vậy phương trình có nghiệm là x=0

Answer 2

mong có thể giúp bạn