`c) |-2x+1| = 3(x-3)`
ĐK : \(\left\{ \begin{array}{l}x\le\dfrac{1}{2}\\x>\dfrac{1}{2}\end{array} \right.\)
Trường hợp 1 : `x \le 1/2`
`-2x + 1 = 3(x-3)`
`⇔ -2x + 1= 3x-9`
`⇔ -2x=3x-10`
`⇔ -5x = -10`
`⇔ x=2\text{(loại)}`
Trường hợp 2 : `x > 1/2`
`-2x + 1 = -3(x-3)`
`⇔ -2x + 1 = -3x + 9`
`⇔ -2x = -3 + 8`
`⇔ x = 8\text{(thỏa mãn)}`
Vậy `S = {8}`
`d) (2x-3)/(x-1) + (-3(x-1))/(x-2) = (3-x^2)/((x-1)(x-2))`
`⇔ (2x-3)/(x-1) - (3(x-1))/(x-2) = (3-x^2)/((x-1)(x-2))`
ĐKXĐ : \(\left\{ \begin{array}{l}x\ne1\\x\ne2\end{array} \right.\)
`⇔ ((2x-3)(x-2))/((x-1)(x-2)) - (3(x-1)^2)/((x-2)(x-1)) = (3-x^2)/((x-1)(x-2))`
`⇒ (2x-3)(x-2) - 3(x-1)^2 = 3-x^2`
`⇔ 2x^2 - 7x + 6 - 3x^2 + 6x - 3 = 3-x^2`
`⇔ -x^2 - x + 3 = 3-x^2`
`⇔ -x^2-x=-x^2`
`⇔ -x = 0`
`⇔ x = 0(TM)`
Vậy `S = {0}`
