\(ĐK:x\ne0;x\ne-1\)
\(\dfrac{2x-5}{x}+\dfrac{4}{x+1}=2\)
\(\Leftrightarrow\dfrac{\left(2x-5\right)\left(x+1\right)+4x}{x\left(x+1\right)}=\dfrac{2x\left(x+1\right)}{x\left(x+1\right)}\)
\(\Leftrightarrow\left(2x-5\right)\left(x+1\right)+4x=2x\left(x+1\right)\)
\(\Leftrightarrow2x^2+2x-5x-5+4x=2x^2+2x\)
\(\Leftrightarrow-5x+4x+2x-2x=5\)
\(\Leftrightarrow-x=5\Leftrightarrow x=-5\left(tm\right)\)
\(\dfrac{2x-5}{x}+\dfrac{4}{x+1}=2\) đkxđ : x khác 0 , x khác -1
\(\Leftrightarrow\dfrac{\left(2x-5\right)\left(x+1\right)}{x\left(x+1\right)}+\dfrac{4x}{x\left(x+1\right)}=\dfrac{2x\left(x+1\right)}{x\left(x+1\right)}\)
\(\Leftrightarrow\dfrac{2x^2+2x-5x-5}{......}+\dfrac{4x}{.....}-\dfrac{2x^2+2x}{....}=0\)
\(\Leftrightarrow2x^2+2x-5x-5+4x-2x^2-2x=0\)
\(\Leftrightarrow-5x-5=0\)
\(\Leftrightarrow x=\left(0+5\right):-5=-1\left(ktm\right)\)
Vậy pt....
