`(2x+3)((3x+8)/(2-7x)+1)=(x-5)((3x+8)/(2-7x)+1)`
`ĐK:x ne 7/2`
`pt<=>((3x+8)/(2-7x)+1)(2x+3-x+5)=0`
`<=>((3x+8)/(2-7x)+1)(x+8)=0`
`<=>` $\left[ \begin{array}{l}x+8=0\\\dfrac{3x+8}{2-7x}=-1\end{array} \right.$
`<=>` $\left[ \begin{array}{l}x=-8\\3x+8=7x-2\end{array} \right.$
`<=>` $\left[ \begin{array}{l}x=-8\\4x=10\end{array} \right.$
`<=>` $\left[ \begin{array}{l}x=-8(TM)\\x=\dfrac{5}{2}(TM)\end{array} \right.$
Vậy `S={-8,5/2}`
ĐK: x \(\ne\dfrac{2}{7}\)
\(\left(2x+3\right)\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)
<=> \(\left(\dfrac{3x+8+2-7x}{2-7x}\right)\left(2x+3-x+5\right)=0\)
<=> \(\left(-4x+10\right)\left(x+8\right)=0\)
<=> \(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-8\end{matrix}\right.\)
