Question

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

Answer 1

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

<=>\(\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x\left(x+1\right)}\)

DKXD x khac 0 =>x khac 0

x+1 khac 0 x khac -1

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

<=>\(\dfrac{\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}-\dfrac{x}{\left(x+1\right)x}=\dfrac{2x-1}{x\left(x+1\right)}\)

<=>\(\dfrac{\left(x-1\right)\left(x+1\right)-x}{x\left(x+1\right)}=\dfrac{2x-1}{x\left(x+1\right)}\)

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

=>\(x^2-1-x=2x-1\)

<=>\(x^2-x-2x=-1+1\)

<=>\(x^2-3x=0\)

<=>x(x-3)=0

<=>x=0(loai)

x-3=0 <=> x=3(thoa man)

Vay x=3 la nghiem cua phuong trinh