Giải hộ mình với ạ a) 2(x-3)=-3(x-1)+7 b) (x^2-1)(x-2)(x-3)=0 c) |5x-3|=x+7

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`a)` `2(x – 3) = -3(x -1) + 7`

`⇔` `2x – 6 = -3x + 3 + 7`

`⇔` `2x – 6 = -3x + 10`

`⇔` `2x – 6 + 3x – 10 = 0`

`⇔` `5x – 16 = 0`

`⇔` `5x = 16`

`⇔` `x = \frac{16}{5}`

`S = {\frac{16}{5}}`

`b)` `(x^2 -1)(x – 2)(x – 3) = 0`

`⇔` `(x – 1)(x + 1)(x – 2)(x – 3) = 0`

`⇔` \(\left[ \begin{array}{l}x – 1 = 0\\x + 1 = 0\\x – 2 = 0\\x – 3 = 0\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}x = 1\\x = -1\\x = 2\\x = 3\end{array} \right.\) 

`S = {-1; 1; 2; 3}`

`c)` `|5x – 3| = x + 7`; ĐKXĐ: `x \ge – 7`

`->` \(\left[ \begin{array}{l}5x – 3 = x +7\\5x – 3 =-(x + 7)\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}5x – 3 – x – 7 = 0\\5x + x = – 7 + 3\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}4x – 10 = 0\\6x = -4\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}4x = 10 \\x = \dfrac{-2}{3}\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}x = \dfrac{5}{2}\\x= \dfrac{-2}{3}\end{array} \right.\) 

`S = {\frac{5}{2}; \frac{-2}{3}}`