\(\dfrac{x}{x^2+4x-5}-\dfrac{2}{x-1}=0\left(x\ne1;-5\right)\\ \Leftrightarrow\dfrac{x}{x^2+5x-x-5}-\dfrac{2}{x-1}=0\\ \Leftrightarrow\dfrac{x}{\left(x+5\right)\left(x-1\right)}-\dfrac{2}{x-1}=0\\ \Leftrightarrow\dfrac{x}{\left(x+5\right)\left(x-1\right)}-\dfrac{2\left(x+5\right)}{\left(x-1\right)\left(x+5\right)}=0\\ \Leftrightarrow x-2\left(x+5\right)=0\\ \Leftrightarrow x-2x-10=0\\ \Leftrightarrow-x-10=0\\ \Leftrightarrow x=-10\left(tm\right)\)
Vậy: ...
đk x khác 1;-5
\(\dfrac{x}{\left(x-1\right)\left(x+5\right)}-\dfrac{2}{x-1}=0\)
\(\Leftrightarrow\dfrac{x-2\left(x+5\right)}{\left(x-1\right)\left(x+5\right)}=0\Leftrightarrow x-2x-10=0\Leftrightarrow-x-10=0\Leftrightarrow x=-10\left(tm\right)\)
