`a)`
`5x(x-3)-x-3=0`
`<=> 5x^2-15x-x-3=0`
`<=> 5x^2-16x-3=0`
`<=> x^2-16/5x-3/5=0`
`<=> x^2-2 . x . 8/5+(8/5)^2-3/5-(8/5)^2=0`
`<=> (x-8/5)^2-79/25=0`
`<=> (x-8/5)^2=79/25`
`<=>` \(\left[ \begin{array}{l}x-\dfrac{8}{5}=\dfrac{\sqrt{79}}{5}\\x-\dfrac{8}{5}=-\dfrac{\sqrt{79}}{5}\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=\dfrac{8+\sqrt{79}}{5}\\x=\dfrac{8-\sqrt{79}}{5}\end{array} \right.\)
Vậy `S={(8+-sqrt(79))/5}`
`b)`
`x^2+3x+x-3=0`
`<=> x^2+4x-3=0`
`<=> x^2+4x+4-7=0`
`<=> (x+2)^2=7`
`<=>` \(\left[ \begin{array}{l}x+2=\sqrt{7}\\x+2=-\sqrt{7}\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=-2+\sqrt{7}\\x=-2-\sqrt{7}\end{array} \right.\)
Vậy `S={-2+-sqrt(7)}`
`c)`
`8x(x+5)-x-5=0`
`<=> 8x(x+5)-(x+5)=0`
`<=> (8x-1)(x+5)=0`
`<=>` \(\left[ \begin{array}{l}8x-1=0\\x+5=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=\dfrac{1}{8}\\x=-5\end{array} \right.\)
Vậy `S={1/8;-5}`
