ĐKXĐ: \(x\ne\frac{2}{3};x\ne-\frac{2}{3}\)
\(\Leftrightarrow\frac{3x+2}{3x-2}-\frac{6}{3x+2}-\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}=0\)
\(\Leftrightarrow\frac{\left(3x+2\right)^2-6\left(3x-2\right)-9x^2}{\left(3x-2\right)\left(3x+2\right)}=0\)
\(\Leftrightarrow9x^2+12x+4-18x+12-9x^2=0\)
\(\Leftrightarrow16-6x=0\)
\(\Leftrightarrow6x=16\)
\(\Leftrightarrow x=\frac{8}{3}\left(TM\right)\)
Vậy \(S=\left\{\frac{8}{3}\right\}\)
\(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}\) (1)
đk: \(x\ne\pm\frac{2}{3}\)
(1)\(\Leftrightarrow\frac{\left(3x+2\right)^2-6\left(3x-2\right)}{\left(3x-2\right)\left(3x+2\right)}=\frac{9x^2}{9x^2-4}\)
\(\Leftrightarrow\frac{9x^2-6x+16}{\left(3x-2\right)\left(3x+2\right)}=\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
\(\Rightarrow9x^2-6x+16=9x^2\)
\(\Leftrightarrow16-6x=0\)
\(\Leftrightarrow x=\frac{8}{3}\)(thỏa mãn đkxđ)
vậy:...................
