\(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{5x-2}{4-x^2}\)
ĐKXĐ: \(x\ne\pm2\)
\(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{5x-2}{4-x^2}\)
\(\Leftrightarrow\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{-5x-2}{x^2-4}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-2\right)}{x^2-4}-\dfrac{x\left(x+2\right)}{x^2-4}=\dfrac{-5x-2}{x^2-4}\)
\(\Leftrightarrow x^2-2x-x+2-x^2-2x=-5x-2\)
\(\Leftrightarrow x^2-2x-x+2-x^2-2x+5x+2=0\)
\(\Leftrightarrow4=0\) (Vô lí)
\(\Rightarrow\) Pt vô \(n_o\)
\(\dfrac{x-1}{x+2}\) - \(\dfrac{x}{x-2}\)= \(\dfrac{5x-2}{4-x^2}\) ĐKXĐ: x \(\ne\) \(\pm\) 2
<=> \(\dfrac{x-1}{x+2}\) + \(\dfrac{x}{2-x}\) = \(\dfrac{5x-2}{\left(x+2\right)\left(2-x\right)}\)
<=> \(\dfrac{\left(x+1\right)\left(2-x\right)}{\left(x+2\right)\left(2-x\right)}\) + \(\dfrac{x\left(x+2\right)}{\left(x+2\right)\left(2-x\right)}\) = \(\dfrac{5x-2}{\left(x+2\right)\left(2-x\right)}\)
<=> \(2x-x^2+2-x\) + \(x^2+2x\) = 5x-2
<=> 3x + 2 = 5x - 2
<=> 3x - 5x = -2 -2
<=> -2x = -4
<=> x = 2 (ko tm ĐKXĐ)
Vậy pt vô nghiệm
