Question

Giải phương trình sau:

\(\dfrac{2}{x-1}+\dfrac{5}{x^2-1}=1\)

Answer 1

\(ĐK:x\ne\pm1\)

\(\Rightarrow\dfrac{2}{x-1}+\dfrac{5}{\left(x-1\right)\left(x+1\right)}=1\)

\(\Leftrightarrow\dfrac{2\left(x+1\right)+5}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(\Leftrightarrow2\left(x+1\right)+5=\left(x-1\right)\left(x+1\right)\)

\(\Leftrightarrow2x+2+5=x^2-1\)

\(\Leftrightarrow x^2-2x-8=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\) (tm)