`8, 13/((x-3)(2x+7)) + 1/(2x+7) = 6/(x^2-9)`
ĐKXĐ : \(\left\{ \begin{array}{l}x\ne3\\x\ne-\dfrac{7}{2}\\x\ne-3\end{array} \right.\)
`⇔ (13(x+3))/((x-3)(2x+7)(x+3)) + (1(x-3)(x+3))/((x-3)(2x+7)(x+3)) = (6(2x+7))/((x-3)(2x+7)(x+3))`
`⇒ 13(x+3) + (x-3)(x+3) = 6(2x+7)`
`⇔ x^2 + 13x + 30 = 12x + 42`
`⇔ x^2 + 13x - 12 = 12x`
`⇔ x^2 + x - 12 = 0`
`⇔ x(x-3)+4(x-3) = 0`
`⇔ (x-3)(x+4) = 0`
`⇔`\(\left[ \begin{array}{l}x-3=0\\x+4=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=3\text{(loại)}\\x=-4(TM)\end{array} \right.\)
Vậy `S = {-4}`
`9, 6/(x-5) + 2/(x-8) = 18/(x^2-13x+40) - 1`
ĐKXĐ : \(\left\{ \begin{array}{l}x\ne5\\x\ne8\end{array} \right.\)
`⇔ (6(x-8))/((x-5)(x-8)) + (2(x-5))/((x-5)(x-8)) = 18/((x-5)(x-8)) - 1(x-5)(x-8)`
`⇒ 6(x-8) + 2(x-5) = 18 - (x-5)(x-8)`
`⇔ 8x - 58 = -x^2 + 13x - 22`
`⇔ -x^2 + 13x - 22 = 8x - 58`
`⇔ -x^2 + 13x + 36 = 8x`
`⇔ -x^2 + 5x + 36 = 0`
`⇔ -(x^2-5x-36) = 0`
`⇔ -(x+4)(x-9) = 0`
`⇔`\(\left[ \begin{array}{l}x+4=0\\x-9=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=-4(TM)\\x=9(TM)\end{array} \right.\)
Vậy `S = {-4,9}`
