\(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x+3}=1-\dfrac{4}{\left(x-1\right)\left(x+3\right)}\)
\(ĐK:x\ne1;x\ne-3\)
\(\Leftrightarrow\dfrac{\left(3x-1\right)\left(x+3\right)-\left(2x+5\right)\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}=\dfrac{\left(x-1\right)\left(x+3\right)-4}{\left(x-1\right)\left(x+3\right)}\)
\(\Leftrightarrow\left(3x-1\right)\left(x+3\right)-\left(2x+5\right)\left(x-1\right)=\left(x-1\right)\left(x+3\right)-4\)
\(\Leftrightarrow3x^2+9x-x-3-\left(2x^2-2x+5x-5\right)=x^2+3x-x-3-4\)
\(\Leftrightarrow3x^2+8x-3-2x^2-3x+5=x^2+2x-7\)
\(\Leftrightarrow3x=-9\)
\(\Leftrightarrow x=-3\left(ktm\right)\)
Vậy pt vô nghiệm
