ĐKXĐ: \(x\ne\pm2\)
\(\frac{1-6x}{x-2}+\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{x^2-4}\)
\(\Leftrightarrow\frac{\left(1-6x\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(9x+4\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{x\left(3x-2\right)+1}{\left(x+2\right)\left(x-2\right)}\)
\(\Leftrightarrow x+2-6x^2-12x+9x^2-18x+4x-8=3x^2-2x+1\)
\(\Leftrightarrow-23x=7\)
\(\Leftrightarrow x=\frac{-7}{23}\left(tm\right)\)
Vậy: \(S=\left\{-\frac{7}{23}\right\}\)
=.= hk tốt!!
\(\text{ĐKXĐ}\: :\: x\ne\pm2\)
\(\frac{1-6x}{x-2}+\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{x^2-4}\)
\(\Leftrightarrow\frac{\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)}{x^2-4}=\frac{x\left(3x-2\right)+1}{x^2-4}\)
Khử mẫu : \(\left(-6x^2-12x+x+2\right)+\left(9x^2-18x+4x-8\right)=3x^2-2x+1\)
\(\Leftrightarrow-23x=7\Leftrightarrow x=\frac{7}{23}\).
