Đáp án:
`a, x^2 - 25 - 8(5-x)= 0`
`<=> x^2 - 25 - 40 + 8x = 0`
`<=> x^2 + 8x - 65=0`
`<=> (x-5)(x+13)= 0`
`<=>`\(\left[ \begin{array}{l}x-5=0\\x+13=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=5\\x=-13\end{array} \right.\)
Vậy `S={-13; 5}`
`b, (2x-1)/3 - (x-5)/4 = (3x+2)/6`
`<=> (8x - 4)/12 - (3x - 15)/12 = (6x + 4)/12`
`=> 8x - 4 - 3x + 15 - 6x - 4=0`
`<=> -x + 7= 0`
`<=> -x=-7`
`<=> x= 7`
Vậy `S={7}`
`c, (x+2)/(x-3) + (x-2)/(x+3)=(2(x^2 + x +7))/(x^2 - 9)` `ĐKXĐ: x≠±3`
`<=> ((x+2)(x+3))/((x-3)(x+3)) + ((x-2)(x-3))/((x-3)(x+3)) = (2x^2 + 2x + 14)/((x-3)(x+3))`
`=>x^2 + 5x + 6 + x^2 - 5x + 6 - 2x^2 - 2x - 14= 0`
`<=> -2x - 2= 0`
`<=> -2x= 2`
`<=> x= -1`
Vậy `S={-1}`
