Giải phương trình sau: (|x-3| - x - 1)( $\frac{x-2}{x-3}$ - $\frac{x-3}{x-2}$ ) = 0

Answer

`(|x - 3| - x - 1) . ({x - 2}/{x - 3} - {x - 3}/{x - 2}) = 0`

Trường hợp `1:`

`|x - 3| - x - 1 = 0`

`<=> |x - 3| = x + 1`

Nhận xét:

`|x - 3| >= 0`

`=> x + 1 >= 0`

`=> x >= -1`

`<=>` $\left[\begin{matrix} x - 3 = x + 1\\ x - 3 = -x - 1\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x - x = 1 + 3\\ x + x = -1 + 3\end{matrix}\right.$

`<=>` $\left[\begin{matrix} 0 = 4 \ \text{(Vô lý)}\\ 2x = 2\end{matrix}\right.$

`<=> 2x = 2`

`<=> x = 2 : 2`

`<=> x = 1` (TMĐK)

Trường hợp `2:`

`{x - 2}/{x - 3} - {x - 3}/{x - 2} = 0` `(Đk: x \ne 3 ; x \ne 2)` 

`<=> {x - 2}/{x - 3} = {x - 3}/{x - 2}`

`<=> (x - 2) . (x - 2) = (x - 3) . (x - 3)`

`<=> (x - 2)^2 = (x - 3)^2`

`<=> x^2 - 4x + 4 = x^2 - 6x + 9`

`<=> x^2 - 4x + 4 - x^2 + 6x - 9 = 0`

`<=> (x^2 - x^2) + (6x - 4x) + (4 - 9) = 0`

`<=> 2x - 5 = 0`

`<=> 2x = 0 + 5`

`<=> 2x = 5`

`<=> x = 5 : 2`

`<=> x = 5/2` (TMĐK)

Vậy `S = {5/2 ; 1}`