a: \(\Leftrightarrow\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)=3x^2-2x+1\)
\(\Leftrightarrow x+2-6x^2-12x+9x^2-18x+4x-8-3x^2+2x-1=0\)
\(\Leftrightarrow-23x=7\)
hay x=-7/23
b: \(\Leftrightarrow1-\dfrac{x}{x-3}=\dfrac{-5x}{\left(x+2\right)\left(x-3\right)}+\dfrac{2}{x+2}\)
\(\Leftrightarrow x^2-x-6-x^2-2x=-5x+2x-6\)
\(\Leftrightarrow-6=-6\)(đúng)
Vậy: S={x|\(x\notin\left\{3;-2\right\}\)}
a) \(\dfrac{1-6x}{x-2}+\dfrac{9x+4}{x+2}=\dfrac{x\left(3x-2\right)+1}{x^2-4}\) ĐKXĐ: \(x\ne\pm2\)
\(\dfrac{\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x\left(3x-2\right)+1}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow x+2-6x^2-12x+9x^2-18x+4x-8=3x^2-2x+1\)
\(\Leftrightarrow-23x=-7\)
\(\Leftrightarrow x=\dfrac{7}{23}\)
b)
