`text{giải phương trình sau :}`
`a,(2x-1)(2+3x)+4-9x^2=0`
`⇔(2x-1)(2+3x)+(2-3x)(2+3x)=0`
`⇔(2+3x)(2-3x+2x-1)=0`
`⇔(2+3x)(1-x)=0`
⇔\(\left[ \begin{array}{l}
2+3x=0\\1-x=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}3x=-2\\x=1\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\frac{-2}{3}\\x=1\end{array} \right.\)
`Vậy S={-2/3;1}`
$b,\dfrac{3}{5x+2} +\dfrac{4}{3x-1} =\dfrac{8x+47}{(5x+2)(3x-1)}$
`MC:(5x+2)(3x-1)`
`ĐKXĐ:xne-2/5 và x ne 1/3`
$⇔\dfrac{3(3x-1)}{(5x+2)(3x-1) } +\dfrac{4(5x+2)}{(3x-1)(5x+2) } =\dfrac{8x+47}{(5x+2)(3x-1)}$
`⇒9x-3+20x+8=8x+47`
`⇔9x+20x-8x=47+3-8`
`⇔21x=42`
`⇔x=42:21`
`⇔x=2(nhận)`
`Vậy S={2}`
